Question 1: Palette 1/42

DPT MCQ

Concept: Distance is the total length of the path covered by a body, regardless of direction. Displacement is the shortest straight-line distance between the initial and final positions, directed from the initial point to the final point. For a straight path, distance equals the magnitude of displacement. For a semicircular path of radius $r$, the path distance is $\pi r$, while the magnitude of displacement is the diameter $2r$.

Ram takes path 1 (straight line) to go from $P$ to $Q$ and Shyam takes path 2 (semicircle) of diameter $100\text{ m}$.
(a) Find the distance travelled by Ram and Shyam.
(b) Find the displacement of Ram and Shyam.

A

(a) Ram $= 100\text{ m}$, Shyam $= 100\text{ m}$; (b) Ram $= 50\pi\text{ m}$, Shyam $= 100\text{ m}$

B

(a) Ram $= 100\text{ m}$, Shyam $= 50\pi\text{ m}$; (b) Ram $= 100\text{ m}$, Shyam $= 100\text{ m}$

C

(a) Ram $= 50\pi\text{ m}$, Shyam $= 100\text{ m}$; (b) Ram $= 100\text{ m}$, Shyam $= 50\pi\text{ m}$

D

(a) Ram $= 100\text{ m}$, Shyam $= 100\pi\text{ m}$; (b) Ram $= 50\text{ m}$, Shyam $= 50\text{ m}$

Question 2: Palette 2/42

DPT MCQ

Concept: Displacement is defined as the change in position of a particle ($\Delta x = x_{\text{final}} - x_{\text{initial}}$). Distance is the actual total length of the path traversed by the body. To find the position at any time $t$, express $x$ in terms of $t$ as $x = (t - 3)^2$.

The position $x$ (in metre) of a particle varies with time $t$ (in second) as $t = \sqrt{x} + 3$. Calculate the:
(a) displacement of the particle from $t = 0$ to $t = 3\text{ s}$
(b) displacement of the particle from $t = 3$ to $t = 6\text{ s}$
(c) distance travelled and displacement from $t = 0$ to $t = 6\text{ s}$

A

(a) $-9\text{ m}$, (b) $+9\text{ m}$, (c) distance $= 18\text{ m}$, displacement $= 0\text{ m}$

B

(a) $+9\text{ m}$, (b) $-9\text{ m}$, (c) distance $= 9\text{ m}$, displacement $= 18\text{ m}$

C

(a) $-9\text{ m}$, (b) $+9\text{ m}$, (c) distance $= 0\text{ m}$, displacement $= 18\text{ m}$

D

(a) $0\text{ m}$, (b) $+9\text{ m}$, (c) distance $= 18\text{ m}$, displacement $= 9\text{ m}$

Question 3: Palette 3/42

DPT MCQ

Concept: Position Vector and Displacement Vector in 3D Coordinate System.
If a particle is located at point $(x, y, z)$, its position vector is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. The displacement vector between initial position $\vec{r}_1$ and final position $\vec{r}_2$ is $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

A particle starts from point $(1, 1, 1)\text{ m}$ and reaches at point $(4, 5, 13)\text{ m}$. Find the initial position vector, final position vector, and displacement vector.

A

Initial: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Final: $\hat{i} + \hat{j} + \hat{k}$, Displacement: $-3\hat{i} - 4\hat{j} - 12\hat{k}$

B

Initial: $\hat{i} + \hat{j} + \hat{k}$, Final: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Displacement: $5\hat{i} + 6\hat{j} + 14\hat{k}$

C

Initial: $\hat{i} + \hat{j} + \hat{k}$, Final: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Displacement: $3\hat{i} + 4\hat{j} + 12\hat{k}$

D

Initial: $3\hat{i} + 4\hat{j} + 12\hat{k}$, Final: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Displacement: $\hat{i} + \hat{j} + \hat{k}$

Question 4: Palette 4/42

DPT MCQ

Concept: Displacement is a vector quantity representing the shortest distance between the initial and final positions. In a three-dimensional Cartesian coordinate system, if a displacement vector is represented as $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude is given by $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2 + z^2}$.

A body moves $6\text{ m}$ north, $8\text{ m}$ east and $10\text{ m}$ vertically upwards. What is its resultant displacement from initial position?

A

$10\text{ m}$

B

$\frac{10}{\sqrt{2}}\text{ m}$

C

$20\text{ m}$

D

$10\sqrt{2}\text{ m}$

Question 5: Palette 5/42

DPT MCQ

Concept: Displacement is the shortest straight-line distance from the initial position to the final position. Since North and East directions are perpendicular ($90^\circ$ to each other), the magnitude of displacement can be calculated using the Pythagorean theorem: $s = \sqrt{x^2 + y^2}$, where $x$ and $y$ are perpendicular components of displacement.

A man goes $10\text{ m}$ towards North, then $20\text{ m}$ towards East. What is his displacement?

A

$25.5\text{ m}$

B

$22.36\text{ m}$

C

$30.0\text{ m}$

D

$25.0\text{ m}$

Question 6: Palette 6/42

DPT MCQ

Concept: The displacement vector between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in a three-dimensional coordinate system is given by $\vec{r} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

An aeroplane flies from $P(-4\text{ m}, -5\text{ m}, +8\text{ m})$ to $Q(7\text{ m}, -2\text{ m}, -3\text{ m})$ in an xyz coordinate system. The displacement position vector of the aeroplane is

A

$11\hat{i} + 3\hat{j} + 11\hat{k}$

B

$11\hat{i} - 3\hat{j} + 11\hat{k}$

C

$11\hat{i} + 3\hat{j} - 11\hat{k}$

D

$11\hat{i} - 3\hat{j} - 11\hat{k}$

Question 7: Palette 7/42

DPT MCQ

Concept: Displacement is the shortest straight-line distance between the initial position ($A$) and final position ($B$). In a quarter-circle path, it forms a right-angled triangle with the radius vectors as legs, so the magnitude of displacement $= \sqrt{R^2 + R^2} = \sqrt{2}R$.
Distance is the total path length traveled along the arc of the circle. The arc length for three-quarters of a full circle ($270^\circ$) from $A$ clockwise to $B$ is $\frac{3}{4} \times 2\pi R = \frac{3\pi R}{2}$.

A body moves in circular path of radius $R$ from $A$ to $B$ as shown. Its displacement and distance covered are

A

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

B

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

C

$\sqrt{2}R, \frac{3\pi R}{2}$

D

None of these

Question 8: Palette 8/42

DPT MCQ

Concept: Distance is the total path length traveled along the semicircular arc, calculated as $d = \pi r$. Displacement is the straight-line distance between the initial and final position, which equals the diameter of the circle, $s = 2r$.

A particle covers half of the circle of radius $r$. Then the displacement and distance of the particle are respectively

A

$2\pi r, 0$

B

$2r, \pi r$

C

$\frac{\pi r}{2}, 2r$

D

$\pi r, r$

Question 9: Palette 9/42

DPT MCQ

Concept: Distance is the total path length traveled by a particle, given by the scalar sum of individual path lengths.
Displacement is the shortest distance between the initial and final positions, represented as a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ with magnitude $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2 + z^2}$.

A particle moves $5\text{ m}$ along east, $6\text{ m}$ along north, and $10\text{ m}$ in the upward direction. Find its distance and displacement.

A

Distance = $21\text{ m}$, Displacement = $15\text{ m}$

B

Distance = $15\text{ m}$, Displacement = $\sqrt{161}\text{ m}$

C

Distance = $21\text{ m}$, Displacement = $\sqrt{161}\text{ m}$

D

Distance = $21\text{ m}$, Displacement = $21\text{ m}$

Question 10: Palette 10/42

DPT MCQ

Concept: Distance is the scalar sum of all individual path lengths covered during motion.
Displacement is the vector sum of individual displacements, representing the net change in position: $\vec{d} = x\hat{i} + y\hat{j}$.

A particle moves $10\text{ m}$ east, $5\text{ m}$ north, $6\text{ m}$ south, $8\text{ m}$ west, $15\text{ m}$ east, and $20\text{ m}$ north. Find its total distance and displacement vector.

A

Distance = $64\text{ m}$, Displacement = $17\hat{i} + 19\hat{j}\text{ m}$

B

Distance = $58\text{ m}$, Displacement = $17\hat{i} + 19\hat{j}\text{ m}$

C

Distance = $64\text{ m}$, Displacement = $19\hat{i} + 17\hat{j}\text{ m}$

D

Distance = $60\text{ m}$, Displacement = $10\hat{i} + 20\hat{j}\text{ m}$

Question 11: Palette 11/42

DPT MCQ

Concept: Average Speed for Equal Distance Intervals
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a particle covers two equal distances $S$ with speeds $v_1$ and $v_2$, the total time taken is the sum of the times taken for each half ($t_1 = \frac{S}{v_1}$ and $t_2 = \frac{S}{v_2}$). The average speed is given by the harmonic mean of the two speeds.

A particle travels half of total distance with speed $v_1$ and next half with speed $v_2$ along a straight line. Find out the average speed of the particle.

A

$\frac{2v_1 v_2}{v_1 + v_2}$

B

$\frac{v_1 + v_2}{2}$

C

$\frac{v_1 v_2}{v_1 + v_2}$

D

$\sqrt{v_1 v_2}$

Question 12: Palette 12/42

DPT MCQ

Concept: When a body covers two equal distances with speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

A car travels the first half of a distance between two places at a speed of $30\text{ km/hr}$ and the second half of the distance at $50\text{ km/hr}$. The average speed of the car for the whole journey is

A

$42.5\text{ km/hr}$

B

$40.0\text{ km/hr}$

C

$37.5\text{ km/hr}$

D

$35.0\text{ km/hr}$

Question 13: Palette 13/42

DPT MCQ

Concept: When a body travels a certain distance $d$ at speed $v_1$ and returns over the same distance $d$ at speed $v_2$, the total distance is $2d$. The average speed is given by the harmonic mean of the two speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

A car travels from $A$ to $B$ at a speed of $20\text{ km h}^{-1}$ and returns at a speed of $30\text{ km h}^{-1}$. The average speed of the car for the whole journey is

A

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

B

$24\text{ km h}^{-1}$

C

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

D

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

Question 14: Palette 14/42

DPT MCQ

Concept: Average Speed for Equal Distances
When a body covers equal distances with different speeds $v_1$ and $v_2$, the average speed for the round trip is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

A car travels from A to B at a speed of $20\text{ km/hr}$ and returns at a speed of $30\text{ km/hr}$. The average speed of the car for the whole journey is

A

$25\text{ km/hr}$

B

$24\text{ km/hr}$

C

$50\text{ km/hr}$

D

$5\text{ km/hr}$

Question 15: Palette 15/42

DPT MCQ

Concept: Average Speed for Equal Distances
When a body covers equal distances with different speeds $v_1$ and $v_2$, the average speed for the total journey is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

A boy walks to his school at a distance of $6\text{ km}$ with constant speed of $2.5\text{ km/hr}$ and walks back with a constant speed of $4\text{ km/hr}$. His average speed for round trip expressed in $\text{km/hour}$, is

A

$\frac{24}{13}$

B

$\frac{40}{13}$

C

$3$

D

$\frac{1}{2}$

Question 16: Palette 16/42

DPT MCQ

Concept: When a body covers two equal distance intervals with speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

A car travels the first half of a distance between two places at a speed of $30\text{ km/hr}$ and the second half of the distance at $50\text{ km/hr}$. The average speed of the car for the whole journey is

A

$42.5\text{ km/hr}$

B

$40.0\text{ km/hr}$

C

$37.5\text{ km/hr}$

D

$35.0\text{ km/hr}$

Question 17: Palette 17/42

DPT MCQ

Concept: Average Speed in a Given Time Interval
$$\text{Average Speed} = \frac{\text{Total Distance Covered}}{\text{Total Time Taken}}$$
To find the average speed over a given time interval, calculate the total distance covered during that exact time period and divide it by the duration of the time interval.

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

A

$5\text{ km/h}$

B

$\frac{25}{4}\text{ km/h}$

C

$\frac{30}{4}\text{ km/h}$

D

$\frac{45}{8}\text{ km/h}$

Question 18: Palette 18/42

DPT MCQ

Concept: Average Speed with Variable Distances and Speeds
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a path is divided into different distance fractions with different constant speeds, calculate the time taken for each section and divide the total distance by the sum of individual time intervals.

One car moving on a straight road covers one third of the distance with $20\text{ km/hr}$ and the rest with $60\text{ km/hr}$. The average speed is

A

$40\text{ km/hr}$

B

$80\text{ km/hr}$

C

$46\frac{2}{3}\text{ km/hr}$

D

$36\text{ km/hr}$

Question 19: Palette 19/42

DPT MCQ

Concept: Average speed is defined as the ratio of total distance traveled to total time taken:
$$v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a journey is divided into segments of known distances and speeds, the total time is calculated by summing the time taken for each individual segment.

If a car covers $\frac{2}{5}\text{th}$ of the total distance with $v_1$ speed and $\frac{3}{5}\text{th}$ distance with $v_2$, then average speed is

A

$\frac{1}{2}\sqrt{v_1 v_2}$

B

$\frac{v_1 + v_2}{2}$

C

$\frac{2v_1 v_2}{v_1 + v_2}$

D

$\frac{5v_1 v_2}{3v_1 + 2v_2}$

Question 20: Palette 20/42

DPT MCQ

Concept: Average Velocity for Equal Time Intervals
$$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
When an object moves with different uniform velocities over equal consecutive time intervals $t$, the average velocity is equal to the arithmetic mean of the individual velocities.

A person travelling on a straight line moves with a uniform velocity $v_1$ for some time and with uniform velocity $v_2$ for the next equal time. The average velocity $v$ is given by

A

$v = \frac{v_1 + v_2}{2}$

B

$v = \sqrt{v_1 v_2}$

C

$v = \frac{2v_1 v_2}{v_1 + v_2}$

D

$v = \frac{v_1 v_2}{v_1 + v_2}$

Question 21: Palette 21/42

DPT MCQ

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
Average speed is a scalar quantity defined as the total path length (distance) divided by the total time taken. Unlike velocity, direction does not affect distance; the distance covered in each segment is simply speed multiplied by time.

A man walks for some time $t$ with velocity $v$ due east. Then he walks for same time $t$ with velocity $v$ due north. The average speed of the man is

A

$2v$

B

$\sqrt{2}v$

C

$v$

D

$\frac{v}{\sqrt{2}}$

Question 22: Palette 22/42

DPT MCQ

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
$$\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}$$
Average speed depends on the total path length traveled, whereas average velocity depends on the net displacement between the initial and final positions. When an object returns to its starting point, its net displacement is zero.

A car travels a distance $d$ on a straight road in two hours and then returns to the starting point in next three hours. Its average speed and average velocity is

A

$\frac{d}{5}, 0$

B

$\frac{2d}{5}, 0$

C

$\frac{5d}{6}, \frac{d}{5}$

D

none of these

Question 23: Palette 23/42

DPT MCQ

Concept: $$\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}$$
Net displacement is the shortest straight-line distance from the initial position to the final position. For perpendicular displacements along two axes ($x$ and $y$), the total displacement magnitude is $\sqrt{x^2 + y^2}$.

A particle moves in the east direction with $15\text{ m/s}$ for $2\text{ s}$ then northwards with $5\text{ m/s}$ for $8\text{ s}$. Average velocity of the particle is

A

$1\text{ m/s}$

B

$5\text{ m/s}$

C

$7\text{ m/s}$

D

$10\text{ m/s}$

Question 24: Palette 24/42

DPT MCQ

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a body covers equal distances with speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

An object travels $10\text{ km}$ at a speed of $100\text{ m/s}$ and another $10\text{ km}$ at $50\text{ m/s}$. The average speed over the whole distance is

A

$75\text{ m/s}$

B

$55\text{ m/s}$

C

$66.7\text{ m/s}$

D

$33.3\text{ m/s}$

Question 25: Palette 25/42

DPT MCQ

Concept: $$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
When a particle moves in equal time intervals $t$ with different velocities $v_1, v_2, v_3, \dots, v_n$ along a straight line in the same direction, the average velocity is equal to the arithmetic mean of the individual velocities:
$$v_{\text{avg}} = \frac{v_1 + v_2 + \dots + v_n}{n}$$

A particle moves in straight line in same direction for $20\text{ sec.}$ with velocity $3\text{ m/s}$ and then moves with velocity $4\text{ m/s}$ for another $20\text{ sec.}$ and finally moves with velocity $5\text{ m/s}$ for next $20\text{ sec.}$. What is the average velocity of the particle?

A

$3\text{ m/s}$

B

$4\text{ m/s}$

C

$5\text{ m/s}$

D

$\text{Zero}$

Question 26: Palette 26/42

DPT MCQ

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a total distance $S$ is divided into equal distance parts traveled at different speeds $v_1, v_2, v_3, \dots, v_n$, the average speed is given by the harmonic mean of the individual speeds:
$$v_{\text{avg}} = \frac{n}{\frac{1}{v_1} + \frac{1}{v_2} + \dots + \frac{1}{v_n}}$$

A body has speed $V$, $2V$ and $3V$ in first $1/3$ part of total travelled distance $S$, second $1/3$ part of $S$ and third $1/3$ part of $S$ respectively. Its average speed will be

A

$V$

B

$2V$

C

$\frac{18}{11}V$

D

$\frac{11}{18}V$

Question 27: Palette 27/42

DPT MCQ

Concept: $$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
When time is divided into equal intervals $t$, and an object moves with uniform velocities $v_1, v_2, \dots, v_n$ during each interval, the average velocity is given by the arithmetic mean of the individual velocities:
$$v_{\text{avg}} = \frac{v_1 + v_2 + \dots + v_n}{n}$$

A body covers one-third of the time with a velocity $v_1$, the second one-third of the time with a velocity $v_2$, and the last one-third of the time with a velocity $v_3$. The average velocity is

A

$\frac{v_1 + v_2 + v_3}{3}$

B

$\frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$

C

$\frac{v_1 v_2 + v_2 v_3 + v_3 v_1}{3}$

D

$\frac{v_1 v_2 v_3}{3}$

Question 28: Palette 28/42

DPT MCQ

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a journey consists of different sections, first calculate the effective average speed for any sub-segment involving equal time intervals:
$$v_{\text{eq time}} = \frac{v_1 + v_2}{2}$$
Then use the formula for equal distance segments to find the total average speed:
$$v_{\text{avg}} = \frac{2 v_{\text{first half}} v_{\text{eq time}}}{v_{\text{first half}} + v_{\text{eq time}}}$$

A particle moving in a straight line covers half the distance with speed of $10\text{ m/s}$. The other half of the distance is covered in two equal time intervals with speed of $4.5\text{ m/s}$ and $7.5\text{ m/s}$ respectively. The average speed of the particle during this motion is

A

$8.0\text{ m/s}$

B

$12.0\text{ m/s}$

C

$10.0\text{ m/s}$

D

$7.5\text{ m/s}$

Question 29: Palette 29/42

DPT MCQ

Concept: $$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
For a multi-stage motion:
1. For equal time intervals, the effective velocity over that part is the arithmetic mean: $v_{\text{eff}} = \frac{v_1 + v_2}{2}$.
2. For two equal distance parts covered with speeds $v_0$ and $v_{\text{eff}}$, the average velocity is given by their harmonic mean: $v_{\text{avg}} = \frac{2 v_0 v_{\text{eff}}}{v_0 + v_{\text{eff}}}$.

A point object traverses half the distance with velocity $v_0$. The remaining part of the distance is covered with velocity $v_1$ for half the time and with velocity $v_2$ for the rest half. The average velocity of the object for the whole journey is

A

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

B

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

C

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

D

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

Question 30: Palette 30/42

DPT MCQ

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a total distance is divided into parts with different constant speeds, the average speed is found by taking the total distance divided by the sum of the time intervals taken for each part.

If a train travels $\frac{1}{4}\text{th}$ of the total distance with speed $v_1$ and the remaining $\frac{3}{4}\text{th}$ distance with speed $v_2$, then its average speed is

A

$\frac{4v_1 v_2}{3v_1 + v_2}$

B

$\frac{v_1 + 3v_2}{4}$

C

$\frac{3v_1 v_2}{v_1 + 3v_2}$

D

$\frac{v_1 v_2}{v_1 + v_2}$

Question 31: Palette 31/42

DPT MCQ

Concept: Instantaneous velocity is the rate of change of position with respect to time, given by the derivative of position $x$ with respect to time $t$, $v = \frac{dx}{dt}$. To find the velocity at a specific moment, differentiate the position function and substitute the given value of time $t$.

The motion of a particle is described by the equation $x = a + bt^2$ where $a = 15 \text{ cm}$ and $b = 3 \text{ cm/s}^2$. Its instantaneous velocity at time $3 \text{ sec}$ will be

A

36 cm/sec

B

18 cm/sec

C

16 cm/sec

D

32 cm/sec

Question 32: Palette 32/42

DPT MCQ

Concept: Average speed is calculated by dividing the total distance travelled by the total time taken for the journey: $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$. The distance for each interval is calculated using the formula $d = v \times t$.

A train has a speed of 60 km/h for the first one hour and 40 km/h for the next half hour. Its average speed in km/h is

A

50

B

53.33

C

48

D

70

Question 33: Palette 33/42

DPT MCQ

Concept: When a total distance is divided into equal distance parts, the average speed is given by the harmonic mean of the individual speeds. Total average speed is calculated using the formula $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$.

A person completes half of his journey with speed $v_1$ and rest half with speed $v_2$. The average speed of the person is

A

$v = \frac{v_1 + v_2}{2}$

B

$v = \frac{2v_1 v_2}{v_1 + v_2}$

C

$v = \frac{v_1 v_2}{v_1 + v_2}$

D

$v = \sqrt{v_1 v_2}$

Question 34: Palette 34/42

DPT MCQ

Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey: $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$. Time for each segment of the journey is calculated using the relation $t = \frac{\text{distance}}{\text{speed}}$.

A car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr. The average speed is

A

40 km/hr

B

80 km/hr

C

$46\frac{2}{3} \text{ km/hr}$

D

36 km/hr

Question 35: Palette 35/42

DPT MCQ

Concept: Acceleration is defined as the second derivative of displacement with respect to time, $a = \frac{d^2 s}{dt^2}$, or the rate of change of velocity with respect to time, $a = \frac{dv}{dt}$ where $v = \frac{ds}{dt}$.

The displacement of a particle, moving in a straight line, is given by $s = 2t^2 + 2t + 4$ where $s$ is in metres and $t$ in seconds. The acceleration of the particle is

A

$2 \text{ m/s}^2$

B

$4 \text{ m/s}^2$

C

$6 \text{ m/s}^2$

D

$8 \text{ m/s}^2$

Question 36: Palette 36/42

DPT MCQ

Concept: Acceleration is defined as the second time derivative of position, $a = \frac{d^2 x}{dt^2}$. To find the time when acceleration becomes zero, differentiate the position function $x(t)$ twice with respect to time $t$ to obtain the acceleration function, set $a(t) = 0$, and solve for $t$.

The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$. The acceleration of the particle will be zero at time $t$ equal to

A

$\frac{a}{b}$

B

$\frac{2a}{3b}$

C

$\frac{a}{3b}$

D

Zero

Question 37: Palette 37/42

DPT MCQ

Concept: Initial velocity is the velocity of the particle at time $t = 0$, obtained by taking the first derivative of displacement with respect to time, $v = \frac{dy}{dt}$. Initial acceleration is the acceleration at $t = 0$, obtained by taking the second derivative of displacement with respect to time, $a = \frac{d^2 y}{dt^2} = \frac{dv}{dt}$.

The displacement of the particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and acceleration are respectively

A

$b, -4d$

B

$-b, 2c$

C

$b, 2c$

D

$2c, -4d$

Question 38: Palette 38/42

DPT MCQ

Concept: Retardation is the magnitude of negative acceleration, $a = -\text{retardation}$. Velocity is the rate of change of position with respect to time, $v = \frac{dx}{dt}$, which can also be written as $v = \frac{1}{\frac{dt}{dx}}$. Acceleration can be expressed in terms of velocity and position using the chain rule: $a = \frac{dv}{dt} = v \frac{dv}{dx}$.

The relation between time $t$ and distance $x$ is $t = \alpha x^2 + \beta x$, where $\alpha$ and $\beta$ are constants. The retardation is ($v$ is the velocity)

A

$2\alpha v^3$

B

$2\beta v^3$

C

$2\alpha\beta v^3$

D

$2\beta^2 v^3$

Question 39: Palette 39/42

DPT MCQ

Concept: When the displacement $x$ of a particle is directly proportional to the square of time $t^2$ (i.e., $x \propto t^2$), we can write $x = Kt^2$ where $K$ is a constant. The velocity is the first time derivative of displacement, $v = \frac{dx}{dt}$, and acceleration is the second time derivative of displacement, $a = \frac{d^2x}{dt^2}$. If acceleration is constant and independent of time, the motion is described as having uniform acceleration.

If displacement of a particle is directly proportional to the square of time, then the particle is moving with

A

Variable acceleration

B

Uniform acceleration

C

Uniform velocity

D

Variable acceleration but uniform velocity

Question 40: Palette 40/42

DPT MCQ

Concept: Average acceleration is defined as the change in velocity divided by the time taken, $\vec{a}_{\text{av}} = \frac{\Delta\vec{v}}{\Delta t} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t}$. Since velocity is a vector quantity, vector subtraction must be used to find $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$. The direction of the average acceleration is along the direction of the change in velocity vector $\Delta\vec{v}$.

A particle is moving eastwards with velocity of $5 \text{ m/s}$. In $10 \text{ sec}$ the velocity changes to $5 \text{ m/s}$ northwards. The average acceleration in this time is

A

Zero

B

$\frac{1}{\sqrt{2}} \text{ m/s}^2$ toward north-west

C

$\frac{1}{\sqrt{2}} \text{ m/s}^2$ toward north-east

D

$\frac{1}{2} \text{ m/s}^2$ toward north-west

Question 41: Palette 41/42

DPT MCQ

Concept: Acceleration is defined as the rate of change of velocity with respect to time, $a = \frac{\Delta v}{t} = \frac{v_2 - v_1}{t}$. Velocity is a vector quantity, so direction must be taken into account when defining initial and final velocities.

A body of mass 10 kg is moving with a constant velocity of $10 \text{ m/s}$. When a constant force acts for $4 \text{ sec}$ on it, it moves with a velocity $2 \text{ m/sec}$ in the opposite direction. The acceleration produced in it is

A

$3 \text{ m/s}^2$

B

$-3 \text{ m/s}^2$

C

$0.3 \text{ m/s}^2$

D

$-0.3 \text{ m/s}^2$

Question 42: Palette 42/42

DPT MCQ

Concept: Displacement is obtained by integrating velocity with respect to time, $x = \int v \, dt$. To find the acceleration, we differentiate velocity with respect to time, $a = \frac{dv}{dt}$. When given the displacement, we first solve for the time $t$ using the position function and then substitute that value of $t$ into the acceleration expression.

A body starts from the origin and moves along the x-axis such that velocity at any instant is given by $(4t^3 - 2t)$, where $t$ is in second and velocity is in m/s. What is the acceleration of the particle, when it is $2 \text{ m}$ from the origin?

A

$28 \text{ m/s}^2$

B

$22 \text{ m/s}^2$

C

$12 \text{ m/s}^2$

D

$10 \text{ m/s}^2$