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Motion in 1D

95 Questions Start Objective Physics Vol-1 Test
All Exams Objective Physics Vol-1 DPT
Q51 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
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
Q52 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
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.
$30 \mathrm{ms}^{-1}$
Q53 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
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
Q54 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
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.
$10\sqrt{6} \mathrm{ms}^{-1}$
Q55 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
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.
$0.03 \mathrm{ms}^{-2}$
Q56 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q57 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q58 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$\frac{2g}{n^{2}}$
Q59 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$\frac{5h}{3}$
Q60 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q61 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$60 \mathrm{ms}^{-1}$
Q62 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$g = \frac{2H}{T_{P}^{2} - T_{Q}^{2}}$
Q63 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q64 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q65 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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 $a = 6t$
Q66 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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 = $\frac{\alpha \sqrt{s}}{2}$
Q67 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q68 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$gt^2 / 2$
Q69 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q70 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q71 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$3/2v_0$
Q72 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$9u^2 / g$
Q73 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$26 \text{ ms}^{-1}$
Q74 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$\left(\frac{u^2}{g^2} - t_1\right)$
Q75 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$t = \sqrt{\frac{t_1}{t_2}}$
Q76 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q77 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q78 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q79 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q80 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q81 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q82 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$37 \text{ ms}^{-2}$
Q83 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q84 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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
Q85 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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.
$v_{0}t + \frac{1}{2} bt^{2}$
Q86 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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 $\text{ms}^{-1}$
Q87 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
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 $\text{ms}^{-1}$
Q88 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: In a position-time (s-t) graph, the slope represents velocity. When the slope is increasing (concave upward), velocity is increasing, which means acceleration is positive.
In the position-time graph shown in Example 3.36, what is the nature of acceleration in region OA? image.png
A.
Zero
B.
Negative
C.
Positive
D.
Cannot be determined
Q89 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: When the position-time graph is a straight line with constant slope, the velocity is constant. If velocity is constant, the acceleration is zero.
For region AB in the position-time graph of Example 3.36, what is the value of acceleration? image.png
A.
Positive constant
B.
Negative constant
C.
Zero
D.
Variable
Q90 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: When the slope of a position-time graph decreases (concave downward), the velocity decreases with time. Decreasing velocity indicates negative acceleration or retardation.
In region BC of the position-time graph, the acceleration is: image.png
A.
Positive
B.
Negative
C.
Zero
D.
Infinite
Q91 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: In region CD, if the slope of the position-time graph is increasing, it indicates that velocity is increasing, which means the particle has positive acceleration.
What can be said about the motion in region CD of the position-time graph? image.png
A.
Velocity is constant
B.
Velocity is decreasing
C.
Velocity is increasing with positive acceleration
D.
Velocity is increasing with negative acceleration
Q92 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: When a position-time graph shows a straight line with constant slope, the object moves with uniform velocity. Uniform velocity means zero acceleration.
For region DE in the position-time graph, which statement is correct?
A.
a > 0, v is increasing
B.
a < 0, v is decreasing
C.
a = 0, v is constant
D.
a is variable, v is variable
Q93 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: In a velocity-time (v-t) graph, the area under the curve with the time axis gives the displacement of the particle. The slope of the v-t graph at any point gives the instantaneous acceleration of the particle.
With the help of the given velocity-time graph, find the (i) displacement in the first three seconds and (ii) acceleration. (Assume the graph is a straight line starting from $v = 30 \text{ m/s}$ at $t = 0$ and reaching $v = 0 \text{ m/s}$ at $t = 3 \text{ s}$). image.png
A.
Displacement = $45 \text{ m}$, Acceleration = $-10 \text{ ms}^{-2}$
B.
Displacement = $90 \text{ m}$, Acceleration = $10 \text{ ms}^{-2}$
C.
Displacement = $45 \text{ m}$, Acceleration = $10 \text{ ms}^{-2}$
D.
Displacement = $90 \text{ m}$, Acceleration = $-10 \text{ ms}^{-2}$
Q94 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: When a vehicle starts from rest, accelerates uniformly, and then decelerates uniformly to rest, its velocity-time graph forms a triangle. The slope of the first part represents acceleration $\alpha$, and the magnitude of the slope of the second part represents deceleration $\beta$. The maximum velocity is the peak of the triangle, and the total distance is the area under the velocity-time graph.
A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$ second, evaluate the maximum velocity reached and the total distance travelled.
A.
$\frac{\alpha \beta t}{\alpha + \beta}$ and $\frac{\alpha \beta t^2}{2(\alpha + \beta)}$
B.
$\frac{\alpha + \beta}{\alpha \beta t}$ and $\frac{\alpha \beta t^2}{2(\alpha + \beta)}$
C.
$\frac{\alpha \beta t}{\alpha + \beta}$ and $\frac{(\alpha + \beta) t^2}{2 \alpha \beta}$
D.
$\frac{(\alpha + \beta) t}{\alpha \beta}$ and $\frac{\alpha \beta t^2}{\alpha + \beta}$
Q95 Objective Physics Vol-1 4. Graphical Analysis MCQ
29 Jul 2026
Concept: The displacement of a particle is equal to the area under its velocity-time graph. By calculating the area of the individual geometric shapes (triangles, rectangles, and trapeziums) formed under the graph for each time interval and summing them up, we can find the total displacement.
A particle moves in a straight line such that its velocity-time graph consists of a triangle from $t=0$ to $t=2$ s reaching $10 \mathrm{m/s}$, a rectangle from $t=2$ to $t=4$ s, a trapezium from $t=4$ to $t=6$ s reaching $20 \mathrm{m/s}$, and a triangle from $t=6$ to $t=8$ s coming to rest. If the initial displacement is zero, what is the total displacement of the particle at the end of 8 seconds? image.png
A.
$60 \mathrm{m}$
B.
$70 \mathrm{m}$
C.
$80 \mathrm{m}$
D.
$100 \mathrm{m}$