Question 1: Palette 1/87

Objective Physics Vol-1 MCQ

Concept: Distance is the total length of the path covered by an object during its motion, regardless of the direction of travel. It is a scalar quantity and is calculated by adding the individual path lengths traversed in each step of the journey.

A scooter is moving along a straight line $AB$ covers a distance of $360\text{ m}$ in $24\text{ s}$ and returns back from $B$ to $C$ and covers $240\text{ m}$ in $18\text{ s}$. Find the total distance travelled by the scooter.

A

$120\text{ m}$

B

$600\text{ m}$

C

$360\text{ m}$

D

$240\text{ m}$

Question 2: Palette 2/87

Objective Physics Vol-1 MCQ

Concept: When a circular wheel rolls without slipping, the total distance covered in $n$ revolutions is equal to $n$ times the circumference of the wheel. The circumference of a circle is given by $\pi d$, where $d$ is the diameter of the wheel.

A wheel completes $2000\text{ revolutions}$ to cover the $9.5\text{ km}$ distance. Find the diameter of the wheel.

A

$1.51\text{ m}$

B

$3.02\text{ m}$

C

$0.75\text{ m}$

D

$2.14\text{ m}$

Question 3: Palette 3/87

Objective Physics Vol-1 MCQ

Concept: Distance is the total path length traversed by a body, regardless of direction. Displacement is the shortest straight-line distance vector from the initial position to the final position, which can be evaluated using vector addition or right-angled triangle geometry (Pythagoras theorem).

A man starts from his home and walks $50\text{ m}$ towards north, then he turns towards east and walks $40\text{ m}$ and then reaches to his office after moving $20\text{ m}$ towards south. What is the total distance covered by the man from his home to office and what is his displacement from his home to office?

A

Distance = $110\text{ m}$, Displacement = $50\text{ m}$

B

Distance = $50\text{ m}$, Displacement = $110\text{ m}$

C

Distance = $110\text{ m}$, Displacement = $70\text{ m}$

D

Distance = $70\text{ m}$, Displacement = $50\text{ m}$

Question 4: Palette 4/87

Objective Physics Vol-1 MCQ

Concept: Distance is the actual length of the path traversed by an object, while displacement is the shortest straight-line distance from its initial position to its final position. For motion along a circular path, distance is calculated as a fraction of the circle's circumference, and displacement is calculated using the straight-line chord length connecting the initial and final positions.

An object covers $1/4\text{th}$ of a circular path of radius $r$. What will be the ratio of the distance and displacement of the object?

A

$\frac{\pi}{2\sqrt{2}}$

B

$\frac{\pi}{\sqrt{2}}$

C

$\frac{2\sqrt{2}}{\pi}$

D

$\frac{\pi}{4}$

Question 5: Palette 5/87

Objective Physics Vol-1 MCQ

Concept: For motion along a semicircular path, the displacement between the initial and final points is equal to the diameter of the path ($2r$). The distance travelled is the actual path length along the semicircle, which is equal to half of the circumference ($\pi r$).

Displacement of a person moving from $X$ to $Y$ along a semicircular path of radius $r$ is $200\text{ m}$. What is the distance travelled by him?

A

$200\text{ m}$

B

$100\pi\text{ m}$

C

$400\text{ m}$

D

$200\pi\text{ m}$

Question 6: Palette 6/87

Objective Physics Vol-1 MCQ

Concept: Distance is the total actual path length covered during motion, which for circular motion depends on the number of completed revolutions multiplied by the circumference ($2\pi r$ or $\pi D$). Displacement is the shortest distance between the initial position and the final position. For an integer number of full rounds, displacement is zero, while for a half round, displacement is equal to the diameter of the track.

An athlete completes one round of a circular track of diameter $200\text{ m}$ in $40\text{ s}$. What will be the distance covered and the displacement at the end of $2\text{ min } 20\text{ s}$?

A

Distance = $2200\text{ m}$, Displacement = $200\text{ m}$

B

Distance = $200\text{ m}$, Displacement = $2200\text{ m}$

C

Distance = $4400\text{ m}$, Displacement = $100\text{ m}$

D

Distance = $2200\text{ m}$, Displacement = $0\text{ m}$

Question 7: Palette 7/87

Objective Physics Vol-1 MCQ

Concept: One-dimensional motion is the motion of a body along a straight line path. In such motion, only one coordinate is required to specify the position of the object at any instant of time.

Which of the following is a one-dimensional motion?

A

Landing of an aircraft

B

Earth revolving around the sun

C

Motion of wheels of moving train

D

Train running on a straight track

Question 8: Palette 8/87

Objective Physics Vol-1 MCQ

Concept: The distance of a point from the origin (or starting point) in three-dimensional space is given by the magnitude of the displacement vector. When an object moves along three mutually perpendicular directions (X, Y, and Z axes), the net direct distance from the initial position is calculated using the 3D distance formula: $d = \sqrt{x^2 + y^2 + z^2}$.

A person moves towards east for $3\text{ m}$, then towards north for $4\text{ m}$ and then moves upwards for $5\text{ m}$. What is his distance now from the starting point?

A

$10\text{ m}$

B

$12\text{ m}$

C

$5\sqrt{2}\text{ m}$

D

$5\text{ m}$

Question 9: Palette 9/87

Objective Physics Vol-1 MCQ

Concept: Distance along a circular arc is equal to the product of the radius of the circle and the angle subtended by the arc at the center measured in radians ($\text{Distance} = R \times \theta$). To convert an angle from degrees to radians, multiply by $\frac{\pi}{180^{\circ}}$.

A particle moves in a circle of radius $R$ from $A$ to $B$ as shown in figure. The distance covered by the object is

A

$\frac{\pi R}{3}$

B

$\frac{\pi R}{2}$

C

$\frac{\pi R}{4}$

D

$\pi R$

Question 10: Palette 10/87

Objective Physics Vol-1 MCQ

Concept: When a wheel rolls forward by half a revolution without slipping, its center moves horizontally by a distance equal to half of its circumference ($\pi R$). Simultaneously, the point initially in contact with the ground moves from the bottom position to the top position of the wheel, resulting in a vertical displacement equal to the diameter of the wheel ($2R$). The net displacement magnitude is the straight-line distance calculated using the Pythagorean theorem: $d = \sqrt{(\pi R)^2 + (2R)^2}$.

A wheel of radius $1\text{ m}$ rolls forward half a revolution on a horizontal ground. The magnitude of displacement of the point of the wheel initially in contact with the ground is

A

$2\pi$

B

$\sqrt{2}\pi$

C

$\sqrt{\pi^2 + 4}$

D

$\pi$

Question 11: Palette 11/87

Objective Physics Vol-1 MCQ

Concept: Displacement ($\Delta x$) on a straight line is defined as the final position ($x_f$) minus the initial position ($x_i$), given by $\Delta x = x_f - x_i$. Displacement is negative when the final position lies to the left of the initial position ($x_f < x_i$).

The three initial and final positions of a man on the x-axis are given as:
(i) $(-8\text{ m}, 7\text{ m})$
(ii) $(7\text{ m}, -3\text{ m})$
(iii) $(-7\text{ m}, 3\text{ m})$
Which pair gives the negative displacement?

A

(i)

B

(ii)

C

(iii)

D

(i) and (iii)

Question 12: Palette 12/87

Objective Physics Vol-1 MCQ

Concept: Displacement is the shortest straight-line distance between the initial and final positions of an object, while distance is the total length of the actual path traversed. Since displacement cannot exceed the total path length, displacement is always less than or equal to distance. Therefore, the ratio of displacement to distance is always less than or equal to one (equal to one for unidirectional motion in a straight line).

The numerical ratio of displacement to the distance for a moving object is always

A

less than one

B

equal to one

C

equal to or less than one

D

equal to or greater than one

Question 13: Palette 13/87

Objective Physics Vol-1 MCQ

Concept: Distance is the actual length of the path traversed by a particle, which for half a revolution of a circular path is equal to half of the circumference ($\pi r$). Displacement is the shortest straight-line distance from the initial position to the final position, which for a half revolution is equal to the diameter of the circular path ($2r$).

A particle moves along a circular path of radius $r$. The distance and displacement of the particle after half a revolution is

A

$0, 2\pi r$

B

$2\pi r, 0$

C

$0, \pi r$

D

$\pi r, 0$

Question 14: Palette 14/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Distance (total path length covered) and Displacement (shortest distance between initial and final position, i.e., final position minus initial position).

A particle starts from the origin, goes along $X$-axis to the point $(20\text{ m}, 0)$ and then returns along the same line to the point $(-20\text{ m}, 0)$. The distance and displacement of the particle during the trip are

A

$40\text{ m}, 0$

B

$40\text{ m}, -20\text{ m}$

C

$60\text{ m}, -20\text{ m}$

D

$40\text{ m}, 20\text{ m}$

Question 15: Palette 15/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed, defined as the total distance traveled divided by the total time taken.

Abdul while driving to school computes the average speed for his trip to be $20\text{ km h}^{-1}$. On his return trip along the same route, there is less traffic and the average speed is $40\text{ km h}^{-1}$. What is the average speed for Abdul's trip?

A

$30.5\text{ km h}^{-1}$

B

$26.67\text{ km h}^{-1}$

C

$24.5\text{ km h}^{-1}$

D

$32.0\text{ km h}^{-1}$

Question 16: Palette 16/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed when the distance is split into equal halves travelled at different constant speeds, calculated using the harmonic mean of the speeds.

A car covers the first half of the distance between two places at a speed of $40\text{ km h}^{-1}$ and second half at $60\text{ km h}^{-1}$. Calculate the average speed of the car.

A

$52\text{ km h}^{-1}$

B

$45\text{ km h}^{-1}$

C

$48\text{ km h}^{-1}$

D

$50\text{ km h}^{-1}$

Question 17: Palette 17/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed for a round trip with two different speeds over equal distances, calculated as the harmonic mean of the speeds.

A car moves from $X$ to $Y$ with a uniform speed $v_u$ and returns to $X$ with a uniform speed $v_d$. Find average speed for this round trip.

A

$\frac{v_u v_d}{v_u + v_d}$

B

$\frac{3v_u v_d}{v_u + v_d}$

C

$\frac{2v_u v_d}{v_u + v_d}$

D

$\frac{v_u + v_d}{2v_u v_d}$

Question 18: Palette 18/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed, defined as the total distance traveled divided by the total time taken for a journey with varying segments of speed and time.

A particle travelled half the distance with a speed $v_0$. The remaining part of the distance was covered with speed $v_1$ for half the time and with speed $v_2$ for the other half of the time. Find the average speed of the particle.

A

$\frac{2v_0(v_1 + v_2)}{v_1 + v_2 + v_0}$

B

$\frac{2v_0(v_1 + v_2)}{v_1 + v_2 + 2v_0}$

C

$\frac{v_0(v_1 + v_2)}{2(v_1 + v_2 + v_0)}$

D

$\frac{v_0(v_1 + v_2)}{v_1 + v_2 + v_0}$

Question 19: Palette 19/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average velocity, defined as the ratio of total displacement to the total time taken for the motion.

In one second, a particle goes from point $A$ to point $B$ moving in a semicircular path as shown in figure. Find the magnitude of average velocity.

A

$1.5\text{ ms}^{-1}$

B

$2.5\text{ ms}^{-1}$

C

$2\text{ ms}^{-1}$

D

$3\text{ ms}^{-1}$

Question 20: Palette 20/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Distance as total path length ($AB + BC + CD$) and Displacement as the shortest straight-line distance from initial to final position using vector components.

A farmer has to go $500\text{ m}$ due north, $400\text{ m}$ due east and $200\text{ m}$ due south to reach his field. If he takes $20\text{ min}$ to reach the field, (i) what distance has he to walk to reach the field? (ii) what is the displacement from his house to the field? (iii) what is the average speed of farmer during the walk? (iv) what is the average velocity of farmer during the walk? [Note: This question specifically focuses on finding the total distance and magnitude of displacement.]

A

Distance: $1100\text{ m}$, Displacement: $500\text{ m}$

B

Distance: $1000\text{ m}$, Displacement: $400\text{ m}$

C

Distance: $1100\text{ m}$, Displacement: $600\text{ m}$

D

Distance: $900\text{ m}$, Displacement: $500\text{ m}$

Question 21: Palette 21/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed defined as total distance divided by total time, and average velocity defined as net displacement divided by total time.

Joseph jogs from one end $A$ to the other end $B$ of a straight $300\text{ m}$ road in $2\text{ min } 50\text{ s}$ and then turns around and jogs $100\text{ m}$ back to point $C$ in another $1\text{ min}$. What are Joseph's average speeds and velocities in jogging (i) from $A$ to $B$ and (ii) from $A$ to $C$?

A

Average speed from $A$ to $B$ is $1.76\text{ ms}^{-1}$, average velocity from $A$ to $B$ is $1.76\text{ ms}^{-1}$; Average speed from $A$ to $C$ is $1.74\text{ ms}^{-1}$, average velocity from $A$ to $C$ is $0.87\text{ ms}^{-1}$

B

Average speed from $A$ to $B$ is $1.50\text{ ms}^{-1}$, average velocity from $A$ to $B$ is $1.50\text{ ms}^{-1}$; Average speed from $A$ to $C$ is $1.20\text{ ms}^{-1}$, average velocity from $A$ to $C$ is $0.50\text{ ms}^{-1}$

C

Average speed from $A$ to $B$ is $2.00\text{ ms}^{-1}$, average velocity from $A$ to $B$ is $2.00\text{ ms}^{-1}$; Average speed from $A$ to $C$ is $1.90\text{ ms}^{-1}$, average velocity from $A$ to $C$ is $1.10\text{ ms}^{-1}$

D

Average speed from $A$ to $B$ is $1.76\text{ ms}^{-1}$, average velocity from $A$ to $B$ is $1.76\text{ ms}^{-1}$; Average speed from $A$ to $C$ is $2.10\text{ ms}^{-1}$, average velocity from $A$ to $C$ is $1.50\text{ ms}^{-1}$

Question 22: Palette 22/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Finding instantaneous position from a given position-time equation, calculating displacement as the difference between final and initial positions, and average velocity as displacement divided by the time interval.

The position of an object moving along $X$-axis is given by $x = 3t - 4t^2 + t^3$, where $x$ is in metres and $t$ in seconds. Find the position of the object at $t = 2\text{ s}$ and $t = 4\text{ s}$. What is the object displacement between $t = 0\text{ s}$ and $t = 4\text{ s}$, and what is its average velocity for the time interval from $t = 2\text{ s}$ to $t = 4\text{ s}$?

A

$x_2 = -2\text{ m}$, $x_4 = 12\text{ m}$, Displacement = $12\text{ m}$, Average Velocity = $7\text{ ms}^{-1}$

B

$x_2 = 2\text{ m}$, $x_4 = 10\text{ m}$, Displacement = $10\text{ m}$, Average Velocity = $5\text{ ms}^{-1}$

C

$x_2 = -4\text{ m}$, $x_4 = 14\text{ m}$, Displacement = $14\text{ m}$, Average Velocity = $8\text{ ms}^{-1}$

D

$x_2 = -2\text{ m}$, $x_4 = 15\text{ m}$, Displacement = $15\text{ m}$, Average Velocity = $9\text{ ms}^{-1}$

Question 23: Palette 23/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed for a journey when time is divided into equal intervals with different constant speeds, calculated as the arithmetic mean of the speeds.

A car has to cover the distance $60\text{ km}$. If half of the total time, it travels with speed $80\text{ km h}^{-1}$ and in rest half time, its speed becomes $40\text{ km h}^{-1}$, the average speed of car will be

A

$120\text{ km h}^{-1}$

B

$80\text{ km h}^{-1}$

C

$60\text{ km h}^{-1}$

D

$180\text{ km h}^{-1}$

Question 24: Palette 24/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed, defined as the total distance traveled divided by the total time taken for the trip.

During the first $18\text{ min}$ of a $60\text{ min}$ trip, a car has an average speed of $11\text{ m min}^{-1}$. What should be the average speed for remaining $42\text{ min}$, so that car is having an average speed of $21\text{ m min}^{-1}$ for the entire trip?

A

$25.3\text{ m min}^{-1}$

B

$29.2\text{ m min}^{-1}$

C

$31\text{ m min}^{-1}$

D

$35.6\text{ m min}^{-1}$

Question 25: Palette 25/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average speed, defined as the total distance traveled divided by the total time taken for the specified time interval.

A man walks on a straight road from his home to a market $2.5\text{ km}$ away with a speed of $5\text{ km h}^{-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5\text{ km h}^{-1}$. The average speed of the man over the interval of time $0$ to $40\text{ min}$ is equal to

A

$5\text{ km h}^{-1}$

B

$\frac{25}{4}\text{ km h}^{-1}$

C

$\frac{30}{4}\text{ km h}^{-1}$

D

$\frac{45}{8}\text{ km h}^{-1}$

Question 26: Palette 26/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Distance, displacement, average speed, and average velocity definitions for linear motion.

A particle is constrained to move on a straight line path. It returns to the starting point after $10\text{ s}$. The total distance covered by the particle during this time is $30\text{ m}$. Which of the following statements about the motion of the particle is true?

A

Displacement of the particle is zero

B

Average speed of the particle is $3\text{ ms}^{-1}$

C

Displacement of the particle is $30\text{ m}$

D

Both (a) and (b)

Question 27: Palette 27/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Uniform speed and distance travelled. The total distance covered by a train to completely cross a bridge is the sum of the length of the train and the length of the bridge.

A $150\text{ m}$ long train is moving with a uniform velocity of $45\text{ km h}^{-1}$. The time taken by the train to cross a bridge of length $850\text{ m}$ is

A

$56\text{ s}$

B

$68\text{ s}$

C

$80\text{ s}$

D

$92\text{ s}$

Question 28: Palette 28/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average velocity, defined as total displacement divided by total time taken.

An insect crawls a distance of $4\text{ m}$ along north in $10\text{ s}$ and then a distance of $3\text{ m}$ along east in $5\text{ s}$. The average velocity of the insect is

A

$\frac{7}{15}\text{ ms}^{-1}$

B

$\frac{1}{5}\text{ ms}^{-1}$

C

$\frac{1}{3}\text{ ms}^{-1}$

D

$\frac{4}{5}\text{ ms}^{-1}$

Question 29: Palette 29/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average velocity defined as the total displacement divided by the total time taken.

A particle traversed $\frac{3}{4}\text{th}$ of the circle of radius $R$ in time $t$. The magnitude of the average velocity of the particle in this time interval is

A

$\frac{\pi R}{t}$

B

$\frac{3\pi R}{2t}$

C

$\frac{R\sqrt{2}}{t}$

D

$\frac{R}{\sqrt{2}t}$

Question 30: Palette 30/87

Objective Physics Vol-1 MCQ

Concept: Kinematics - Average velocity along a semicircular path, defined as the displacement (diameter of the semicircle) divided by the time taken to cover the semicircular arc.

A boy is running over a circular track with uniform speed of $10\text{ ms}^{-1}$. What is the average velocity for movement of boy along semicircle (in $\text{ms}^{-1}$)?

A

$\frac{10}{\pi}$

B

$\frac{40}{\pi}$

C

$10$

D

$\frac{20}{\pi}$

Question 31: Palette 31/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for uniformly accelerated motion

A car starts from rest, attains a velocity of $18 \mathrm{kmh}^{-1}$ with an acceleration of $0.5 \mathrm{ms}^{-2}$, travels 4 km with this uniform velocity and then comes to halt with a uniform deceleration of $0.4 \mathrm{ms}^{-2}$. Calculate the total time of travel of the car.

A

812.5 s

B

802.5 s

C

822.5 s

D

832.5 s

Question 32: Palette 32/87

Objective Physics Vol-1 MCQ

Concept: Vector nature of acceleration and its dependence on velocity

Acceleration of a particle changes when

A

direction of velocity changes

B

magnitude of velocity changes

C

Both (a) and (b)

D

speed changes

Question 33: Palette 33/87

Objective Physics Vol-1 MCQ

Concept: Relationship between acceleration, velocity, and speed

If a particle moves with an acceleration, then which of the following can remain constant?

A

Both speed and velocity

B

Neither speed nor velocity

C

Only the velocity

D

Only the speed

Question 34: Palette 34/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations relating average velocity, distance, time, and acceleration

The average velocity of a body moving with uniform acceleration travelling a distance of $3.06 \mathrm{m}$ is $0.34 \mathrm{ms}^{-1}$. If the change in velocity of the body is $0.18 \mathrm{ms}^{-1}$, then during this time, its uniform acceleration is

A

$0.01 \mathrm{ms}^{-2}$

B

$0.02 \mathrm{ms}^{-2}$

C

$0.03 \mathrm{ms}^{-2}$

D

$0.04 \mathrm{ms}^{-2}$

Question 35: Palette 35/87

Objective Physics Vol-1 MCQ

Concept: Calculation of retardation using the first equation of motion

A car travelling with a velocity of $80 \mathrm{km/h}$ slowed down to $44 \mathrm{km/h}$ in $15 \mathrm{s}$. The retardation is

A

$0.67 \mathrm{ms}^{-2}$

B

$1 \mathrm{ms}^{-2}$

C

$1.25 \mathrm{ms}^{-2}$

D

$1.5 \mathrm{ms}^{-2}$

Question 36: Palette 36/87

Objective Physics Vol-1 MCQ

Concept: Acceleration in straight-line motion versus curved-path motion at constant speed

An object is moving along the path OABO with constant speed, then

A

the acceleration of the object while moving along the path OABO is zero

B

the acceleration of the object along the path OA and BO is zero

C

there must be some acceleration along the path AB

D

Both (b) and (c)

Question 37: Palette 37/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for uniformly accelerated motion

Two cars start off a race with velocities $2 \mathrm{ms}^{-1}$ and $4 \mathrm{ms}^{-1}$ and travel in a straight line with uniform accelerations $2 \mathrm{ms}^{-2}$ and $1 \mathrm{ms}^{-2}$, respectively. What is the length of the path, if they reach the final point at the same time?

A

12 m

B

24 m

C

36 m

D

48 m

Question 38: Palette 38/87

Objective Physics Vol-1 MCQ

Concept: Third equation of motion and calculation of retardation

A car was moving at a rate of $18 \mathrm{kmh}^{-1}$. When the brakes were applied, it comes to rest at a distance of 100 m. Calculate the retardation produced by the brakes.

A

$0.125 \mathrm{ms}^{-2}$

B

$0.250 \mathrm{ms}^{-2}$

C

$0.500 \mathrm{ms}^{-2}$

D

$1.000 \mathrm{ms}^{-2}$

Question 39: Palette 39/87

Objective Physics Vol-1 MCQ

Concept: Stopping distance using the third equation of motion

Two cars are travelling towards each other on a straight road at velocities $10 \mathrm{ms}^{-1}$ and $12 \mathrm{ms}^{-1}$, respectively. When they are 150 m apart, both the drivers apply their brakes and each car decelerates at $2 \mathrm{ms}^{-2}$ until it stops. How far apart will they be when both of them come to rest?

A

61 m

B

75 m

C

89 m

D

100 m

Question 40: Palette 40/87

Objective Physics Vol-1 MCQ

Concept: Relative motion and time of flight for constant velocity

A train travelling at $20 \mathrm{kmh}^{-1}$ is approaching a platform. A bird is sitting on a pole on the platform. When the train is at a distance of 2 km from the pole, brakes are applied which produce a uniform deceleration in it. At that instant, the bird flies towards the train at $60 \mathrm{kmh}^{-1}$ and after touching the nearest point on the train flies back to the pole and then flies towards the train and continues repeating itself. Calculate how much distance the bird covers before the train stops?

A

6 km

B

12 km

C

18 km

D

24 km

Question 41: Palette 41/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for uniformly accelerated motion applied in segments

A particle starts with an initial velocity and passes successively over the two halves of a given distance with accelerations $a_1$ and $a_2$, respectively. The final velocity is the same as if the whole distance is covered with a uniform acceleration of

A

$\frac{a_1 - a_2}{2}$

B

$\frac{2a_1 a_2}{a_1 + a_2}$

C

$\frac{a_1 + a_2}{2}$

D

$\sqrt{a_1 a_2}$

Question 42: Palette 42/87

Objective Physics Vol-1 MCQ

Concept: Equations of motion for two bodies starting from rest

In a car race, car A takes a time $t$ less than car B at the finish point and passes the finishing point with speed $v$ more than that of the car B. Assuming that both the cars start from rest and travel with constant accelerations $a_1$ and $a_2$ respectively, the correct relation is

A

$v = \sqrt{a_1 a_2}t$

B

$v = (a_1 + a_2)t$

C

$v = \sqrt{\frac{a_1}{a_2}}t$

D

$v = \frac{a_1 a_2}{t}$

Question 43: Palette 43/87

Objective Physics Vol-1 MCQ

Concept: Distance travelled in the nth second of uniformly accelerated motion

A body starting from rest has an acceleration of $4 \mathrm{ms}^{-2}$. The distance travelled by it in the 5th second is

A

20 m

B

18 m

C

16 m

D

24 m

Question 44: Palette 44/87

Objective Physics Vol-1 MCQ

Concept: Ratio of displacements in successive time intervals for a body starting from rest

A particle starts from rest and moves under constant acceleration in a straight line. The ratio of displacements in successive seconds is

A

1:2:3:...

B

1:4:9:...

C

2:4:6:...

D

1:3:5:...

Question 45: Palette 45/87

Objective Physics Vol-1 MCQ

Concept: First equation of motion for uniformly accelerated motion

Velocity of a body moving along a straight line with uniform acceleration reduces by $\frac{3}{4}$ of its initial velocity in time $t_{0}$. The total time of motion of the body till its velocity becomes zero is

A

$\frac{4}{3}t_{0}$

B

$\frac{3}{2}t_{0}$

C

$\frac{5}{3}t_{0}$

D

$\frac{8}{3}t_{0}$

Question 46: Palette 46/87

Objective Physics Vol-1 MCQ

Concept: Second equation of motion for displacement

The displacement of a body in 8 s starting from rest with an acceleration of $20 \mathrm{cms}^{-2}$ is

A

$64 \mathrm{m}$

B

$64 \mathrm{cm}$

C

$640 \mathrm{cm}$

D

$0.064 \mathrm{m}$

Question 47: Palette 47/87

Objective Physics Vol-1 MCQ

Concept: Integration of velocity to find displacement and distance

The motion of a particle is described by the equation $v = at$. The distance travelled by the particle in the first 4 s is

A

$4a$

B

$12a$

C

$6a$

D

$8a$

Question 48: Palette 48/87

Objective Physics Vol-1 MCQ

Concept: Second equation of motion for calculating time

A particle starts with a velocity of $2 \mathrm{ms}^{-1}$ and moves in a straight line with a retardation of $0.1 \mathrm{ms}^{-2}$. The first time at which the particle is 15 m from the starting point is

A

10 s

B

20 s

C

30 s

D

40 s

Question 49: Palette 49/87

Objective Physics Vol-1 MCQ

Concept: Calculation of total distance in different phases of motion

A particle starts from rest, accelerates at $2 \mathrm{ms}^{-2}$ for 10 s and then moves with constant speed of $20 \mathrm{ms}^{-1}$ for 30 s and then decelerates at $4 \mathrm{ms}^{-2}$ till it stops after next 5 s. What is the distance travelled by it?

A

750 m

B

800 m

C

700 m

D

850 m

Question 50: Palette 50/87

Objective Physics Vol-1 MCQ

Concept: Equation of motion for uniform velocity and uniform acceleration

A body is moving with uniform velocity of $8 \mathrm{ms}^{-1}$. When the body just crossed another body, the second one starts and moves with uniform acceleration of $4 \mathrm{ms}^{-2}$. The time after which two bodies meet, will be

A

2 s

B

4 s

C

6 s

D

8 s

Question 51: Palette 51/87

Objective Physics Vol-1 MCQ

Concept: Second equation of motion for displacement

Two bodies A and B start from rest from the same point with a uniform acceleration of $2 \mathrm{ms}^{-2}$. If B starts one second later, then the two bodies are separated at the end of the next second, by

A

1 m

B

2 m

C

3 m

D

4 m

Question 52: Palette 52/87

Objective Physics Vol-1 MCQ

Concept: Calculation of average velocity using total displacement and total time

A train accelerating uniformly from rest attains a maximum speed of $40 \mathrm{ms}^{-1}$ in 20 s. It travels at this speed for 20 s and is brought to rest by uniform retardation in further 40 s. What is the average velocity during this period?

A

$\frac{80}{3} \mathrm{ms}^{-1}$

B

$40 \mathrm{ms}^{-1}$

C

$25 \mathrm{ms}^{-1}$

D

$30 \mathrm{ms}^{-1}$

Question 53: Palette 53/87

Objective Physics Vol-1 MCQ

Concept: Relation between average speed, maximum speed, and distances in different phases of motion

A particle starts from rest and traverses a distance $l$ with uniform acceleration, then moves uniformly over a further distance $2l$ and finally comes to rest after moving a further distance $3l$ under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its ways is

A

1/5

B

2/5

C

3/5

D

4/5

Question 54: Palette 54/87

Objective Physics Vol-1 MCQ

Concept: Third equation of motion applied to find velocity at the midpoint

A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 \mathrm{ms}^{-1}$ and $30 \mathrm{ms}^{-1}$, respectively. The speed of the body at mid-point of $A$ and $B$ is

A

$25 \mathrm{ms}^{-1}$

B

$25.5 \mathrm{ms}^{-1}$

C

$24 \mathrm{ms}^{-1}$

D

$10\sqrt{6} \mathrm{ms}^{-1}$

Question 55: Palette 55/87

Objective Physics Vol-1 MCQ

Concept: Distance travelled in the nth second of uniformly accelerated motion

If a body starts from rest and travels 120 cm in the 6th second, then what is the acceleration?

A

$0.20 \mathrm{ms}^{-2}$

B

$0.027 \mathrm{ms}^{-2}$

C

$0.218 \mathrm{ms}^{-2}$

D

$0.03 \mathrm{ms}^{-2}$

Question 56: Palette 56/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for motion under gravity

A ball is thrown upwards from the top of a tower 40 m high with a velocity of 10 m/s. Find the time when it strikes the ground. (Take, $g = 10 \mathrm{m/s}^{2}$)

A

4 s

B

2 s

C

5 s

D

3 s

Question 57: Palette 57/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity in two phases

A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10 \mathrm{ms}^{-2}$. The fuel is finished in 1 min and it continues to move up. What is the maximum height reached by the rocket? (Take, $g = 10 \mathrm{ms}^{-2}$)

A

18 km

B

54 km

C

36 km

D

72 km

Question 58: Palette 58/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity and time of ascent

A juggler throws balls into air. He throws one ball whenever the previous one is at its highest point. How high does the balls rise, if he throws $n$ balls each second? (Acceleration due to gravity is $g$)

A

$\frac{g}{n^{2}}$

B

$\frac{g}{2n^{2}}$

C

$\frac{g}{4n^{2}}$

D

$\frac{2g}{n^{2}}$

Question 59: Palette 59/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for motion under gravity

From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance $h$ below A, its velocity is double of what it was at a height $h$ above A. What is the greatest height attained by the stone?

A

$\frac{h}{3}$

B

$\frac{2h}{3}$

C

$\frac{4h}{3}$

D

$\frac{5h}{3}$

Question 60: Palette 60/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for motion under gravity

A ball is thrown vertically upwards with a velocity of $20 \mathrm{ms}^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is 25 m from the ground. How long will it take before the ball hits the ground? (Take, $g = 10 \mathrm{ms}^{-2}$)

A

5 s

B

2 s

C

3 s

D

4 s

Question 61: Palette 61/87

Objective Physics Vol-1 MCQ

Concept: Kinematic equations for motion under gravity and quadratic equation

A ball is thrown upwards from the ground with an initial speed $u$. The ball is at a height of 80 m at two times, for the time interval of 6 s. Find the value of $u$. (Take, $g = 10 \mathrm{ms}^{-2}$)

A

$30 \mathrm{ms}^{-1}$

B

$40 \mathrm{ms}^{-1}$

C

$50 \mathrm{ms}^{-1}$

D

$60 \mathrm{ms}^{-1}$

Question 62: Palette 62/87

Objective Physics Vol-1 MCQ

Concept: Time of flight and kinematic equations for motion under gravity

A particle is thrown vertically upwards from the surface of the earth. Let $T_{P}$ be the time taken by the particle to travel from a point P above the earth to its highest point and back to the point P. Similarly, let $T_{Q}$ be the time taken by the particle to travel from another point Q above the earth to its highest point and back to the same point Q. If the distance between the points P and Q is $H$, find the expression for acceleration due to gravity in terms of $T_{P}$, $T_{Q}$ and $H$.

A

$g = \frac{4H}{T_{P}^{2} - T_{Q}^{2}}$

B

$g = \frac{8H}{T_{P}^{2} - T_{Q}^{2}}$

C

$g = \frac{8H}{T_{P}^{2} + T_{Q}^{2}}$

D

$g = \frac{2H}{T_{P}^{2} - T_{Q}^{2}}$

Question 63: Palette 63/87

Objective Physics Vol-1 MCQ

Concept: Free fall under gravity at equal time intervals

From the top of a building 16 m high, water drops are falling at equal intervals of time such that when the first drop reaches the ground, the fifth drop just starts. Find the distance between the successive drops at that instant.

A

7 m, 5 m, 3 m, 1 m

B

5 m, 4 m, 3 m, 2 m

C

6 m, 4 m, 3 m, 1 m

D

8 m, 5 m, 3 m, 2 m

Question 64: Palette 64/87

Objective Physics Vol-1 MCQ

Concept: Relative motion of two freely falling bodies

A ball is dropped from the top of a tower. After 2 s, another ball is thrown vertically downwards with a speed of $40 \mathrm{ms}^{-1}$. After how much time and at what distance below the top of the tower do the balls meet? (Take, $g = 10 \mathrm{ms}^{-2}$)

A

2 s, 20 m

B

3 s, 45 m

C

4 s, 80 m

D

5 s, 125 m

Question 65: Palette 65/87

Objective Physics Vol-1 MCQ

Concept: Non-uniformly accelerated motion and calculus-based kinematics. When acceleration is not constant, the basic equations of velocity and acceleration are derived using differentiation and integration. Velocity is the rate of change of displacement ($v = \frac{ds}{dt}$), and acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$). Displacement can be found by integrating the velocity equation with respect to time, applying the given boundary conditions.

The velocity-time equation of a particle moving in a straight line is given by $v = 10 + 2t + 3t^2$ (in SI units). If the displacement of the particle is 20 m at time $t = 0$, what is the displacement of the particle at time $t = 1$ s and its corresponding acceleration-time equation?

A

Displacement = 32 m, Acceleration $a = 2 + 6t$

B

Displacement = 20 m, Acceleration $a = 2 + 3t$

C

Displacement = 32 m, Acceleration $a = 10 + 2t$

D

Displacement = 12 m, Acceleration $a = 6t$

Question 66: Palette 66/87

Objective Physics Vol-1 MCQ

Concept: Non-uniformly accelerated motion and calculus-based kinematics. When velocity is given as a function of displacement ($v = f(x)$), we can use the relation $v = \frac{dx}{dt}$ to separate variables and integrate to find the displacement-time or velocity-time relationship. The mean velocity is defined as the total displacement divided by the total time taken.

The velocity of a particle moving in the positive direction of the X-axis varies as $v = \alpha \sqrt{x}$, where $\alpha$ is a positive constant. Assuming that at moment $t = 0$, the particle was located at the point $x = 0$. What is the time dependence of the velocity of the particle and the mean velocity of the particle averaged over the time that the particle takes to cover the first $s$ metres of the path?

A

$v = \alpha^2 t$, Mean velocity = $\alpha \sqrt{s}$

B

$v = \frac{1}{2} \alpha^2 t$, Mean velocity = $\frac{\alpha \sqrt{s}}{2}$

C

$v = \frac{1}{2} \alpha^2 t$, Mean velocity = $\alpha \sqrt{s}$

D

$v = \alpha^2 t$, Mean velocity = $\frac{\alpha \sqrt{s}}{2}$

Question 67: Palette 67/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. When an object is thrown vertically upwards, the time of ascent is equal to the time of descent. The total time of flight is given by $T = \frac{2u}{g}$.

If a stone is thrown up with a velocity of $9.8 \text{ ms}^{-1}$, then how much time will it take to come back?

A

1 s

B

2 s

C

3 s

D

4 s

Question 68: Palette 68/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The distance covered in the last $t$ seconds of ascent is equal to the distance covered in the first $t$ seconds of free fall from the maximum height, where the initial velocity is zero.

If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ second of its ascent is

A

$ut - (gt^2 / 2)$

B

$(u + gt)t$

C

$ut$

D

$gt^2 / 2$

Question 69: Palette 69/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The time taken to reach the maximum height is the time of ascent. If the next ball is thrown when the first ball's velocity is zero, the time of ascent is equal to the interval between throws. The maximum height can be found using $h = \frac{1}{2}gt^2$.

A person throws balls into air after every second. The next ball is thrown when the velocity of the first ball is zero. How high do the balls rise above his hand?

A

2 m

B

5 m

C

8 m

D

10 m

Question 70: Palette 70/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. Using the third equation of motion $v^2 - u^2 = 2as$, we can relate the velocities at different heights. At maximum height, the final velocity is zero.

A particle is thrown vertically upwards. Its velocity at half of the height is $10 \text{ ms}^{-1}$. Then, the maximum height attained by it is (Take, $g = 10 \text{ ms}^{-2}$)

A

16 m

B

10 m

C

20 m

D

40 m

Question 71: Palette 71/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The maximum height attained by a body thrown vertically upwards is directly proportional to the square of its initial velocity, given by $h = \frac{v_0^2}{2g}$.

When a ball is thrown up vertically with velocity $v_0$, it reaches a maximum height of $h$. If one wishes to triple the maximum height, then the ball should be thrown with velocity,

A

$\sqrt{3}v_0$

B

$3v_0$

C

$9v_0$

D

$3/2v_0$

Question 72: Palette 72/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. Using the third equation of motion $v^2 - u^2 = 2as$, taking the downward direction as positive to relate the initial velocity, final velocity, and the height of the tower.

A stone thrown upward with a speed $u$ from the top of the tower reaches the ground with a speed $3u$. The height of the tower is

A

$3u^2 / g$

B

$4u^2 / g$

C

$6u^2 / g$

D

$9u^2 / g$

Question 73: Palette 73/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The time interval between passing a certain height twice is the time it takes to go from that height to the maximum height and back. The velocity at that height can be found using the time of flight for that specific segment, assuming $g = 10 \text{ ms}^{-2}$ for calculation.

A body thrown vertically up from the ground passes the height of 10.2 m twice in an interval of 10 s. What was its initial velocity?

A

$52 \text{ ms}^{-1}$

B

$61 \text{ ms}^{-1}$

C

$45 \text{ ms}^{-1}$

D

$26 \text{ ms}^{-1}$

Question 74: Palette 74/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The total time of flight to reach a certain height and return is $\frac{2u}{g}$. The time to reach the point is $t_1$, and the time to return from the peak is determined by the symmetry of the motion.

A body is projected with a velocity $u$. It passes through a certain point above the ground after $t_1$ second. The time interval after which the body passes through the same point during the return journey is

A

$\left(\frac{u}{g} - t_1^2\right)$

B

$2\left(\frac{u}{g} - t_1\right)$

C

$\left(\frac{u}{g} - t_1\right)$

D

$\left(\frac{u^2}{g^2} - t_1\right)$

Question 75: Palette 75/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. Using the second equation of motion $s = ut + \frac{1}{2}at^2$ for the three different cases (upward, downward, and free fall) from the same height to find the relationship between the times.

A body is thrown vertically upwards from the top A of tower. It reaches the ground in $t_1$ second. If it is thrown vertically downwards from A with the same speed, it reaches the ground in $t_2$ second. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by

A

$t = \frac{t_1 + t_2}{2}$

B

$t = \frac{t_1 - t_2}{2}$

C

$t = \sqrt{t_1 t_2}$

D

$t = \sqrt{\frac{t_1}{t_2}}$

Question 76: Palette 76/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity and relative motion. First, find the velocity and height of the balloon when the ball is released. Then, treat the ball as a projectile thrown upwards with that initial velocity from that height, under the influence of gravity.

A man in a balloon rising vertically with an acceleration of $4.9 \text{ ms}^{-2}$ releases a ball 2 s after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (Take, $g = 9.8 \text{ ms}^{-2}$)

A

14.7 m

B

19.6 m

C

9.8 m

D

24.5 m

Question 77: Palette 77/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The distance covered in the $n$-th second of a freely falling body is $S_n = \frac{1}{2}g(2n - 1)$, and the distance covered in the first $t$ seconds is $S = \frac{1}{2}gt^2$.

A stone falls freely under gravity. The total distance covered by it in the last second of its journey equals the distance covered by it in first 3 s of its motion. The time for which stone remains in air, is

A

5 s

B

12 s

C

15 s

D

8 s

Question 78: Palette 78/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The distance travelled in time $t$ is $S = \frac{1}{2}gt^2$. We can find the distance travelled in successive 2-second intervals and then determine their ratio.

A body falls from a height $h = 200 \text{ m}$. The ratio of distance travelled in each $2 \text{ s}$, during $t = 0$ to $t = 6 \text{ s}$ of the journey is

A

1:4:9

B

1:2:4

C

1:3:5

D

1:2:3

Question 79: Palette 79/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The time taken by a body to fall freely from a height $H$ is given by $t = \sqrt{\frac{2H}{g}}$.

A ball is released from height $h$ and another from $2h$. The ratio of time taken by the two balls to reach the ground is

A

$1:\sqrt{2}$

B

$\sqrt{2}:1$

C

2:1

D

1:2

Question 80: Palette 80/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The total distance is $h = \frac{1}{2}gt^2$ and the distance covered in the last ($t$-th) second is $S_t = \frac{1}{2}g(2t - 1)$. Equating these based on the given condition allows us to find the total time $t$ and subsequently the height $h$.

A particle is dropped under gravity from rest from a height $h$ ($g = 9.8 \text{ ms}^{-2}$) and it travels a distance $9h / 25$ in the last second, the height $h$ is

A

100 m

B

1225 m

C

145 m

D

167.5 m

Question 81: Palette 81/87

Objective Physics Vol-1 MCQ

Concept: Motion under gravity. The distance covered in the first second of free fall is $x = \frac{1}{2}g(1)^2$, and the distance covered in the $n$-th second is $S_n = \frac{1}{2}g(2n - 1)$.

A body dropped from the top of a tower covers a distance $7x$ in the last second of its journey, where $x$ is the distance covered in first second. How much time does it take to reach the ground?

A

3 s

B

4 s

C

5 s

D

6 s

Question 82: Palette 82/87

Objective Physics Vol-1 MCQ

Concept: Average acceleration is the change in velocity divided by the time interval. Velocity is the first derivative of displacement with respect to time, and acceleration is the derivative of velocity. If acceleration is constant, the average acceleration equals the instantaneous acceleration.

The displacement (in metre) of a particle moving along X-axis is given by $x = 18t + 5t^{2}$. The average acceleration during the interval $t_{1} = 2s$ and $t_{2} = 4s$ is

A

$13 \text{ ms}^{-2}$

B

$10 \text{ ms}^{-2}$

C

$27 \text{ ms}^{-2}$

D

$37 \text{ ms}^{-2}$

Question 83: Palette 83/87

Objective Physics Vol-1 MCQ

Concept: Distance is the total path length covered by the particle. To find the total distance, we must determine if the particle changes direction by finding when its velocity is zero. The total distance is the sum of the absolute displacements for each interval where the direction remains constant.

The displacement of a particle moving in a straight line is described by the relation $s = 6 + 12t - 2t^{2}$. Here, s is in metre and t is in second. The distance covered by particle in first 5 s is

A

20 m

B

32 m

C

24 m

D

26 m

Question 84: Palette 84/87

Objective Physics Vol-1 MCQ

Concept: Initial velocity and initial acceleration are found by differentiating the displacement equation with respect to time to get velocity and acceleration, and then evaluating these expressions at $t = 0$.

The displacement of a particle moving in a straight line depends on time as $x = \alpha t^{3} + \beta t^{2} + \gamma t + \delta$. The ratio of initial acceleration to its initial velocity depends on

A

$\alpha$ and $\gamma$ only

B

$\beta$ and $\gamma$ only

C

$\alpha$ and $\beta$ only

D

$\alpha$ only

Question 85: Palette 85/87

Objective Physics Vol-1 MCQ

Concept: When acceleration is given as a function of time, velocity is found by integrating acceleration with respect to time, applying the initial velocity condition. Displacement is then found by integrating the velocity equation with respect to time, applying the initial position condition.

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity $v_{0}$. The distance travelled by the particle in time t will be

A

$v_{0}t + \frac{1}{6} bt^{3}$

B

$v_{0}t + \frac{1}{3} bt^{3}$

C

$v_{0}t + \frac{1}{3} bt^{2}$

D

$v_{0}t + \frac{1}{2} bt^{2}$

Question 86: Palette 86/87

Objective Physics Vol-1 MCQ

Concept: Velocity is the integral of acceleration with respect to time. By integrating the given acceleration function and applying the initial velocity condition, we can find the velocity at any specific time.

The acceleration $a$ (in $\text{ms}^{-2}$), of a particle is given by $a = 3t^2 + 2t + 2$, where $t$ is the time. If the particle starts out with a velocity $v = 2 \text{ ms}^{-1}$ at $t = 0$, then the velocity at the end of 2 s is

A

12 $\text{ms}^{-1}$

B

14 $\text{ms}^{-1}$

C

16 $\text{ms}^{-1}$

D

18 $\text{ms}^{-1}$

Question 87: Palette 87/87

Objective Physics Vol-1 MCQ

Concept: Velocity is the first derivative of displacement. To find the minimum velocity, we differentiate the velocity function with respect to time (which gives acceleration), set it to zero to find the critical point, and verify it is a minimum using the second derivative test.

A particle is moving such that $s = t^{3} - 6t^{2} + 18t + 9$, where s is in metre and t is in second. The minimum velocity attained by the particle is

A

29 $\text{ms}^{-1}$

B

5 $\text{ms}^{-1}$

C

6 $\text{ms}^{-1}$

D

12 $\text{ms}^{-1}$