Kinematics-1D
207 Questions
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Q101
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: Displacement is given by the net area under the velocity-time graph, considering signs: $\text{Displacement} = \int v \, dt$. Speed is the magnitude of velocity: $\text{Speed} = \vert{}v\vert{}$. A change in the sign of velocity indicates a change in the direction of motion.
The figure shows the velocity of a particle plotted against time $t$ from the image given below
A.
The displacement of the particle is zero
B.
The particle changes its direction of motion at some point
C.
The initial and final speeds of the particle are same
D.
All of the above statements are correct
Q102
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: Average velocity is defined as the total displacement divided by the total time taken: $\text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}}$. The displacement over a time interval corresponds to the net area bounded by the velocity-time graph and the time axis, taking area above the axis as positive and below as negative.
The $v-t$ plot of a moving object is shown in the figure from the image given below. The average velocity of the object during the first 10 seconds is
A.
0
B.
$2.5 \text{ ms}^{-1}$
C.
$5 \text{ ms}^{-1}$
D.
$2 \text{ ms}^{-1}$
Q103
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: For a physically possible motion, velocity must be a single-valued function of time $t$. At any given instant of time $t$, a particle can have only one unique velocity value. If a graph shows multiple values of velocity for a single point in time, or if time moves backwards, that graph represents a physical impossibility.
Which of the following velocity time graphs is possible from the image given below
A.
B.
C.
D.
Q104
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: The total distance travelled during uniform acceleration, constant speed, and uniform deceleration is the sum of displacements in each stage: $s_{\text{total}} = s_1 + s_2 + s_3$. The formulas involved are $v = u + at$ and $s = ut + \frac{1}{2}at^2$ for accelerated motion, $s = vt$ for constant speed motion, and $v^2 = u^2 + 2as$ for decelerated motion.
A particle starts from rest, accelerates at $2 \text{ m/s}^2$ for $10\text{ s}$ and then goes for constant speed for $30\text{ s}$ and then decelerates at $4 \text{ m/s}^2$ till it stops. What is the distance travelled by it
A.
750 m
B.
800 m
C.
700 m
D.
850 m
Q105
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: Acceleration is the slope or rate of change of velocity with respect to time, $a = \frac{dv}{dt}$. On a velocity-time graph, a straight line with a constant negative slope corresponds to a constant negative acceleration, a horizontal line with zero slope corresponds to zero acceleration, and a straight line with a constant positive slope corresponds to a constant positive acceleration.
The graph below shows the velocity versus time graph for a body from the image given below. Which of the following graphs represents the corresponding acceleration versus time graphs
A.
B.
C.
D.
Q106
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: Velocity is obtained by integrating acceleration with respect to time, $v(t) = \int a \, dt + v_0$. When acceleration $a$ is positive and constant, velocity $v$ increases linearly with time (positive slope). When acceleration $a$ is zero, velocity $v$ remains constant (zero slope).
The acceleration-time graph for a body is shown in the following graph from the image given below. Which of the following graphs would probably represent the velocity of the body plotted against time
A.
B.
C.
D.
Q107
Error less
v-t Graph
MCQ
29 Jul 2026
Concept: Velocity is the rate of change of displacement with respect to time, given by the derivative $v = \frac{dx}{dt}$. When $x$ is a quadratic function of time, $v(t)$ is a linear function of time represented by a straight line graph with a specific y-intercept and slope.
A particle is moving in such a way that its displacement is related with time by the equation $x = (10 - 4t + 6t^2) \text{ m}$. The diagram showing variation of velocity of particle with time is from the image given below
A.
B.
C.
D.
Q108
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: The distance covered by a moving body is equal to the total area enclosed under its velocity-time graph.
Formula: $S = \text{Area under } v - t \text{ graph} = \frac{1}{2} \times (a + b) \times h$
In the velocity-time graph from the image given below, the distance travelled by the body in metres is
Formula: $S = \text{Area under } v - t \text{ graph} = \frac{1}{2} \times (a + b) \times h$
A.
200
B.
250
C.
300
D.
400
Q109
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: The distance covered by an object moving along a straight line is equal to the area under its velocity-time graph.
Formula for area of a trapezium: $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Formula for area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
For the velocity-time graph shown from the image given below, the distance covered by the body in the last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds?
Formula for area of a trapezium: $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
Formula for area of a triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
A.
1/2
B.
1/4
C.
1/3
D.
2/3
Q110
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: Displacement is the vector sum of the areas under the velocity-time graph taking direction into account, whereas distance is the total magnitude sum of all areas under the graph.
Formula for displacement: $\text{Displacement} = A_1 + A_2 + A_3$
Formula for distance: $\text{Distance} = \vert{}A_1\vert{} + \vert{}A_2\vert{} + \vert{}A_3\vert{}$
The velocity-time graph of a body moving in a straight line is shown from the image gven below. The displacement and distance travelled by the body in 6 s are respectively
Formula for displacement: $\text{Displacement} = A_1 + A_2 + A_3$
Formula for distance: $\text{Distance} = \vert{}A_1\vert{} + \vert{}A_2\vert{} + \vert{}A_3\vert{}$
A.
8 m, 16 m
B.
16 m, 8 m
C.
16 m, 16 m
D.
8 m, 8 m
Q111
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: For a body projected vertically upward under gravity, the acceleration is constant and directed downwards ($a = -g$).
The velocity varies linearly with time according to the equation of motion: $v = u - gt$.
During the upward motion, velocity decreases linearly to zero in the positive region. During the downward motion, velocity increases linearly in magnitude in the negative direction.
A ball is thrown vertically upward. Which of the following graphs from the image given below represents the velocity-time graph of the ball during its flight when air resistance is neglected?
The velocity varies linearly with time according to the equation of motion: $v = u - gt$.
During the upward motion, velocity decreases linearly to zero in the positive region. During the downward motion, velocity increases linearly in magnitude in the negative direction.
A.
B.
C.
D.
Q112
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: For motion under gravity, the relation between velocity $v$ and height $h$ is given by $v^2 = u^2 + 2g(d - h)$, which represents a parabolic curve.
When the ball drops, $h$ decreases from $d$ to $0$ while downward velocity increases. Upon bouncing, the velocity reverses direction (becomes positive) and its magnitude drops, then $h$ increases back to $d/2$ as velocity decreases to zero.
A ball is dropped vertically from a height $d$ above the ground. It hits the ground and bounces up vertically to a height $d/2$. Neglecting subsequent motion and air resistance, its velocity $v$ varies with the height $h$ above the ground as shown from the image given below:
When the ball drops, $h$ decreases from $d$ to $0$ while downward velocity increases. Upon bouncing, the velocity reverses direction (becomes positive) and its magnitude drops, then $h$ increases back to $d/2$ as velocity decreases to zero.
A.
B.
C.
D.
Q113
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: Velocity is obtained by integrating acceleration with respect to time: $v = \int a \, dt$.
If acceleration increases linearly with time ($a = kt$), velocity increases quadratically ($v \propto t^2$), forming a parabolic curve. When acceleration suddenly drops to zero ($a = 0$), the velocity remains constant.
The acceleration-time graph of a body is shown from the image given below. The most probable velocity-time graph of the body is
If acceleration increases linearly with time ($a = kt$), velocity increases quadratically ($v \propto t^2$), forming a parabolic curve. When acceleration suddenly drops to zero ($a = 0$), the velocity remains constant.
A.
B.
C.
D.
Q114
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: A particle or body moving along a path cannot have two different values of velocity at a single instant of time.
Formula: A valid single-valued function requires $v = v(t)$ to yield a unique value for every given $t$.
Which of the following velocity-time graphs is not possible from the image given below?
Formula: A valid single-valued function requires $v = v(t)$ to yield a unique value for every given $t$.
A.
B.
C.
D.
Q115
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: According to Newton's second law, force is directly proportional to acceleration ($F = ma$). Thus, the ratio of forces in two time intervals equals the ratio of accelerations, which is given by the slope of the velocity-time graph ($\tan \theta$).
Formula: $\text{Force ratio} = \frac{F_1}{F_2} = \frac{a_1}{a_2} = \frac{\tan \theta_1}{\tan \theta_2}$
For a certain body, the velocity-time graph is shown from the image given below. The ratio of applied forces for intervals AB and BC is
Formula: $\text{Force ratio} = \frac{F_1}{F_2} = \frac{a_1}{a_2} = \frac{\tan \theta_1}{\tan \theta_2}$
A.
$+ \frac{1}{\sqrt{3}}$
B.
$- \frac{1}{\sqrt{3}}$
C.
$+ \sqrt{3}$
D.
$- \frac{1}{3}$
Q116
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: The slope of a velocity-time graph represents the acceleration of the moving body.
Formula: $a = \frac{dv}{dt} = \tan \theta$
Velocity-time graphs of two cars which start from rest at the same time, are shown from the image given below. Graph shows, that
Formula: $a = \frac{dv}{dt} = \tan \theta$
A.
Initial velocity of A is greater than the initial velocity of B
B.
Acceleration in A is increasing at lesser rate than in B
C.
Acceleration in A is greater than in B
D.
Acceleration in B is greater than in A
Q117
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: When a ball falls freely under gravity, its downward velocity increases linearly with time due to constant downward acceleration ($g$). Upon bouncing off a hard surface, its velocity instantaneously reverses direction (becomes positive or upward) with a slightly reduced magnitude, and then decreases linearly back to zero at peak height.
Formula: $v = u - gt$
Which one of the following graphs represents the velocity of a steel ball which falls from a height on to a marble floor from the image given below? (Here $v$ represents the velocity of the particle and $t$ the time)
Formula: $v = u - gt$
A.
B.
C.
D.
Q118
Error less
v-t Graph
MCQ
31 Jul 2026
Concept: Acceleration is given by the slope of the velocity-time graph.
Formula: $a = \frac{v_2 - v_1}{t_2 - t_1}$
The adjoining curve represents the velocity-time graph of a particle, its acceleration values along OA, AB and BC in $m/s^2$ from the image given below are respectively
Formula: $a = \frac{v_2 - v_1}{t_2 - t_1}$
A.
1, 0, -0.5
B.
1, 0, 0.5
C.
1, 1, 0.5
D.
1, 0.5, 0
Q119
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: When two bodies starting from the same point meet after time $t$, their total displacements in that time interval must be equal ($s_A = s_B$).
Formula for displacement with initial rest ($u=0$) under uniform acceleration: $s_A = \frac{1}{2}at^2$
Formula for displacement with constant velocity: $s_B = vt$
A body A moves with a uniform acceleration $a$ and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:
Formula for displacement with initial rest ($u=0$) under uniform acceleration: $s_A = \frac{1}{2}at^2$
Formula for displacement with constant velocity: $s_B = vt$
A.
$\frac{2v}{a}$
B.
$\frac{v}{a}$
C.
$\frac{v}{2a}$
D.
$\sqrt{\frac{v}{2a}}$
Q120
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: For the student to catch the bus, the distance covered by the student must equal the initial gap plus the displacement of the bus.
Distance travelled by student: $s_{\text{student}} = ut$
Displacement of bus: $s_{\text{bus}} = \frac{1}{2}at^2$
Condition for meeting: $ut = 50 + \frac{1}{2}at^2 \implies \frac{1}{2}at^2 - ut + 50 = 0$
For real time $t$, the discriminant of this quadratic equation must be greater than or equal to zero ($D \ge 0$).
A student is standing at a distance of $50\text{ metres}$ from the bus. As soon as the bus starts its motion with an acceleration of $1\text{ ms}^{-2}$, the student starts running towards the bus with a uniform velocity $u$. Assuming the motion to be along a straight road, the minimum value of $u$, so that the student is able to catch the bus is
Distance travelled by student: $s_{\text{student}} = ut$
Displacement of bus: $s_{\text{bus}} = \frac{1}{2}at^2$
Condition for meeting: $ut = 50 + \frac{1}{2}at^2 \implies \frac{1}{2}at^2 - ut + 50 = 0$
For real time $t$, the discriminant of this quadratic equation must be greater than or equal to zero ($D \ge 0$).
A.
$8\text{ ms}^{-1}$
B.
$12\text{ ms}^{-1}$
C.
$10\text{ ms}^{-1}$
D.
$5\text{ ms}^{-1}$
Q121
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: Stopping distance ($s$) under uniform retarding force/acceleration ($a$) is given by $v^2 = u^2 + 2as$.
Since final velocity $v = 0$, $0 = u^2 - 2as \implies s = \frac{u^2}{2a}$.
For a constant retarding acceleration, stopping distance is directly proportional to the square of initial velocity ($s \propto u^2$).
A car, moving with a speed of $50\text{ km/hr}$, can be stopped by brakes after at least $6\text{ m}$. If the same car is moving at a speed of $100\text{ km/hr}$, the minimum stopping distance is
Since final velocity $v = 0$, $0 = u^2 - 2as \implies s = \frac{u^2}{2a}$.
For a constant retarding acceleration, stopping distance is directly proportional to the square of initial velocity ($s \propto u^2$).
A.
$6\text{ m}$
B.
$12\text{ m}$
C.
$18\text{ m}$
D.
$24\text{ m}$
Q122
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: Retardation is the uniform negative acceleration experienced by a body when its speed decreases over a given distance.
Using the third equation of motion: $v^2 = u^2 - 2as$, where $u$ is initial velocity, $v$ is final velocity, $a$ is retardation, and $s$ is displacement/thickness.
Formula for retardation: $a = \frac{u^2 - v^2}{2s}$
The velocity of a bullet is reduced from $200\text{ m/s}$ to $100\text{ m/s}$ while travelling through a wooden block of thickness $10\text{ cm}$. The retardation, assuming it to be uniform, will be
Using the third equation of motion: $v^2 = u^2 - 2as$, where $u$ is initial velocity, $v$ is final velocity, $a$ is retardation, and $s$ is displacement/thickness.
Formula for retardation: $a = \frac{u^2 - v^2}{2s}$
A.
$10 \times 10^4\text{ m/s}^2$
B.
$12 \times 10^4\text{ m/s}^2$
C.
$13.5 \times 10^4\text{ m/s}^2$
D.
$15 \times 10^4\text{ m/s}^2$
Q123
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: Distance travelled by a body in the $n\text{th}$ second of its motion starting from rest ($u=0$) under uniform acceleration $a$ is given by $S_n = \frac{a}{2}(2n - 1)$.
Since body B starts $2\text{ seconds}$ after body A, the $5\text{th}$ second for body A corresponds to the $3\text{rd}$ second for body B ($n_B = 5 - 2 = 3$).
A body A starts from rest with an acceleration $a_1$. After $2\text{ seconds}$, another body B starts from rest with an acceleration $a_2$. If they travel equal distances in the $5\text{th}$ second, after the start of A, then the ratio $a_1 : a_2$ is equal to
Since body B starts $2\text{ seconds}$ after body A, the $5\text{th}$ second for body A corresponds to the $3\text{rd}$ second for body B ($n_B = 5 - 2 = 3$).
A.
$5:9$
B.
$5:7$
C.
$9:5$
D.
$9:7$
Q124
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: Time taken during motion is given by: $t = \frac{\text{Distance}}{\text{Average velocity}}$
Uniform acceleration is the rate of change of velocity: $a = \frac{\text{Change in velocity}}{\text{Time}}$
The average velocity of a body moving with uniform acceleration travelling a distance of $3.06\text{ m}$ is $0.34\text{ ms}^{-1}$. If the change in velocity of the body is $0.18\text{ ms}^{-1}$ during this time, its uniform acceleration is
Uniform acceleration is the rate of change of velocity: $a = \frac{\text{Change in velocity}}{\text{Time}}$
A.
$0.01\text{ ms}^{-2}$
B.
$0.02\text{ ms}^{-2}$
C.
$0.03\text{ ms}^{-2}$
D.
$0.04\text{ ms}^{-2}$
Q125
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: Using the second equation of motion under constant acceleration: $s = ut + \frac{1}{2}at^2$.
Formulate equations for total distance travelled from start ($t = 0$) at two different time intervals ($t = 5\text{ s}$ and $t = 8\text{ s}$) to solve for initial velocity $u$ and acceleration $a$.
Then calculate the total distance travelled in $t = 10\text{ s}$ and subtract the distance travelled in the first $8\text{ s}$.
A particle travels $10\text{ m}$ in first $5\text{ sec}$ and $10\text{ m}$ in next $3\text{ sec}$. Assuming constant acceleration what is the distance travelled in next $2\text{ sec}$
Formulate equations for total distance travelled from start ($t = 0$) at two different time intervals ($t = 5\text{ s}$ and $t = 8\text{ s}$) to solve for initial velocity $u$ and acceleration $a$.
Then calculate the total distance travelled in $t = 10\text{ s}$ and subtract the distance travelled in the first $8\text{ s}$.
A.
$8.3\text{ m}$
B.
$9.3\text{ m}$
C.
$10.3\text{ m}$
D.
None of above
Q126
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: Displacement starting from rest ($u = 0$) under constant acceleration $a$ for time $t$ is given by $S = \frac{1}{2}at^2$.
The distances covered in equal consecutive intervals of time starting from rest follow Galileo's law of odd numbers, i.e., $S_1 : S_2 : S_3 = 1 : 3 : 5$.
A body travels for $15\text{ sec}$ starting from rest with constant acceleration. If it travels distances $S_1$, $S_2$ and $S_3$ in the first five seconds, second five seconds and next five seconds respectively the relation between $S_1$, $S_2$ and $S_3$ is
The distances covered in equal consecutive intervals of time starting from rest follow Galileo's law of odd numbers, i.e., $S_1 : S_2 : S_3 = 1 : 3 : 5$.
A.
$S_1 = S_2 = S_3$
B.
$5S_1 = 3S_2 = S_3$
C.
$S_1 = \frac{1}{3}S_2 = \frac{1}{5}S_3$
D.
$S_1 = \frac{1}{5}S_2 = \frac{1}{3}S_3$
Q127
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: The distance travelled by a body in the $n\text{th}$ second of its motion with initial velocity $u$ and uniform acceleration $a$ is given by $S_n = u + \frac{1}{2}a(2n - 1)$.
If a body having initial velocity zero is moving with uniform acceleration $8\text{ m/sec}^2$, the distance travelled by it in fifth second will be
A.
$36\text{ metres}$
B.
$40\text{ metres}$
C.
$100\text{ metres}$
D.
Zero
Q128
DPT
Uniform Acceleration
MCQ
04 Aug 2026
Concept: According to Newton's second law, $F = ma$, which implies that for a constant force $F$, acceleration is inversely proportional to mass ($a \propto \frac{1}{m}$).
Formula: $\frac{a_2}{a_1} = \frac{m_1}{m_2}$
The engine of a car produces acceleration $4\text{ m/s}^2$ in the car, if this car pulls another car of same mass, what will be the acceleration produced
Formula: $\frac{a_2}{a_1} = \frac{m_1}{m_2}$
A.
$8\text{ m/s}^2$
B.
$2\text{ m/s}^2$
C.
$4\text{ m/s}^2$
D.
$\frac{1}{2}\text{ m/s}^2$
Q129
Revision
MCQ
15 Aug 2026
Concept: When a wheel of radius $R$ rolls forward without slipping by half a revolution, the horizontal displacement of the point in contact with the ground is equal to half the circumference of the wheel, $x = \pi R$. Simultaneously, the point rotates to the top of the wheel, undergoing a vertical displacement equal to the diameter of the wheel, $y = 2R$. The net displacement vector is $\vec{S} = (\pi R)\hat{i} + (2R)\hat{j}$, and its magnitude is given by the Pythagorean theorem:
$S = \sqrt{x^2 + y^2} = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$
The displacement of the point of the wheel initially in contact with the ground, when the wheel rolls forward half a revolution will be (radius of the wheel is $R$) from the image given below
$S = \sqrt{x^2 + y^2} = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$
A.
$\frac{R}{\sqrt{\pi^2 + 4}}$
B.
$R\sqrt{\pi^2 + 4}$
C.
$2\pi R$
D.
$\pi R$
Q130
Revision
MCQ
15 Aug 2026
Concept: Average speed is defined as the total distance travelled divided by the total time taken for the journey:
$v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{t_1 + t_2}$
Where the time taken for each section is calculated using $t = \frac{\text{distance}}{\text{speed}}$.
If a car covers $2/5\text{th}$ of the total distance with $v_1$ speed and $3/5\text{th}$ distance with $v_2$ then average speed is from the image given below
$v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{t_1 + t_2}$
Where the time taken for each section is calculated using $t = \frac{\text{distance}}{\text{speed}}$.
A.
$\frac{1}{2}\sqrt{v_1 v_2}$
B.
$\frac{v_1 + v_2}{2}$
C.
$\frac{2 v_1 v_2}{v_1 + v_2}$
D.
$\frac{5 v_1 v_2}{3 v_1 + 2 v_2}$
Q131
Revision
MCQ
15 Aug 2026
Concept: Average velocity is defined as the total displacement divided by total time:
$v_{av} = \frac{\Delta r}{\Delta t}$
When an object starts and ends its journey at the same position, its net displacement is zero ($\Delta r = 0$), making its average velocity zero. However, speed is the rate of total distance covered over time, which increases as long as the object continues moving.
A car accelerated from initial position and then returned at initial point, then
$v_{av} = \frac{\Delta r}{\Delta t}$
When an object starts and ends its journey at the same position, its net displacement is zero ($\Delta r = 0$), making its average velocity zero. However, speed is the rate of total distance covered over time, which increases as long as the object continues moving.
A.
Velocity is zero but speed increases
B.
Speed is zero but velocity increases
C.
Both speed and velocity increase
D.
Both speed and velocity decrease
Q132
Revision
MCQ
15 Aug 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken for a given interval of time:
$v_{av} = \frac{\text{Total distance}}{\text{Total time}}$
Time required to travel a distance $d$ at speed $v$ is given by $t = \frac{d}{v}$.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min. is equal to
$v_{av} = \frac{\text{Total distance}}{\text{Total time}}$
Time required to travel a distance $d$ at speed $v$ is given by $t = \frac{d}{v}$.
A.
5 km/h
B.
$\frac{25}{4}$ km/h
C.
$\frac{30}{4}$ km/h
D.
$\frac{45}{8}$ km/h
Q133
Revision
MCQ
15 Aug 2026
Concept: Instantaneous velocity is the time derivative of displacement, given by $v = \frac{dx}{dt}$.
To find the displacement when velocity is zero, first express $x$ as a function of $t$, differentiate it to find the velocity expression $v(t)$, set $v(t) = 0$ to determine the time $t$, and then substitute that time back into the displacement equation $x(t)$.
The relation $3t = \sqrt{3x} + 6$ describes the displacement of a particle in one direction where $x$ is in metres and $t$ in sec. The displacement, when velocity is zero, is
To find the displacement when velocity is zero, first express $x$ as a function of $t$, differentiate it to find the velocity expression $v(t)$, set $v(t) = 0$ to determine the time $t$, and then substitute that time back into the displacement equation $x(t)$.
A.
24 metres
B.
12 metres
C.
5 metres
D.
Zero
Q134
Revision
MCQ
15 Aug 2026
Concept: Instantaneous velocity is defined as the rate of change of position vector/displacement with respect to time. Mathematically, it is the first derivative of the position function $x(t)$ with respect to time $t$:
$v = \frac{dx}{dt}$
The motion of a particle is described by the equation $x = a + bt^2$ where $a = 15\text{ cm}$ and $b = 3\text{ cm}$. Its instantaneous velocity at time $3\text{ sec}$ will be
$v = \frac{dx}{dt}$
A.
$36\text{ cm/sec}$
B.
$18\text{ cm/sec}$
C.
$16\text{ cm/sec}$
D.
$32\text{ cm/sec}$
Q135
Revision
MCQ
15 Aug 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey:
$v_{av} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$
Distance covered in a given interval is calculated using $d = v \times t$.
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
$v_{av} = \frac{\text{Total distance travelled}}{\text{Total time taken}}$
Distance covered in a given interval is calculated using $d = v \times t$.
A.
50
B.
53.33
C.
48
D.
70
Q136
Revision
MCQ
15 Aug 2026
Concept: Average speed for a journey split into equal distance intervals is calculated using the total distance divided by the total time taken:
$v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d_1 + d_2}{\frac{d_1}{v_1} + \frac{d_2}{v_2}}$
When the total distance is divided into two equal parts ($d_1 = d_2 = \frac{d}{2}$), the average speed is the harmonic mean of the two speeds:
$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$
A person completes half of its his journey with speed $v_1$ and rest half with speed $v_2$. The average speed of the person is
$v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d_1 + d_2}{\frac{d_1}{v_1} + \frac{d_2}{v_2}}$
When the total distance is divided into two equal parts ($d_1 = d_2 = \frac{d}{2}$), the average speed is the harmonic mean of the two speeds:
$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$
A.
$v = \frac{v_1 + v_2}{2}$
B.
$v = \frac{2 v_1 v_2}{v_1 + v_2}$
C.
$v = \frac{v_1 v_2}{v_1 + v_2}$
D.
$v = \sqrt{v_1 v_2}$
Q137
Revision
MCQ
15 Aug 2026
Concept: Average speed is defined as total distance covered divided by total time taken:
$v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{t_1 + t_2}$
Where the time for each segment is given by $t = \frac{\text{distance}}{\text{speed}}$.
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
$v_{av} = \frac{\text{Total distance}}{\text{Total time}} = \frac{x}{t_1 + t_2}$
Where the time for each segment is given by $t = \frac{\text{distance}}{\text{speed}}$.
A.
40 km/hr
B.
80 km/hr
C.
$46\frac{2}{3}$ km/hr
D.
36 km/hr
Q138
Revision
MCQ
15 Aug 2026
Concept: Acceleration is defined as the time rate of change of velocity, which in turn is the time rate of change of displacement.
Velocity $v = \frac{ds}{dt}$
Acceleration $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$
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
Velocity $v = \frac{ds}{dt}$
Acceleration $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$
A.
$2\text{ m/s}^2$
B.
$4\text{ m/s}^2$
C.
$6\text{ m/s}^2$
D.
$8\text{ m/s}^2$
Q139
Revision
MCQ
15 Aug 2026
Concept: Acceleration is defined as the second derivative of the position function $x(t)$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
To find the time when acceleration is zero, 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
$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
To find the time when acceleration is zero, set $a(t) = 0$ and solve for $t$.
A.
$\frac{a}{b}$
B.
$\frac{2a}{3b}$
C.
$\frac{a}{3b}$
D.
Zero
Q140
Revision
MCQ
15 Aug 2026
Concept: Instantaneous velocity is the time derivative of displacement ($v = \frac{dy}{dt}$) and instantaneous acceleration is the time derivative of velocity ($a = \frac{dv}{dt}$).
Initial velocity and initial acceleration refer to the values of velocity and acceleration evaluated at time $t = 0$.
The displacement of the particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and acceleration are respectively
Initial velocity and initial acceleration refer to the values of velocity and acceleration evaluated at time $t = 0$.
A.
$b, -4d$
B.
$-b, 2c$
C.
$b, 2c$
D.
$2c, -4d$
Q141
Revision
MCQ
15 Aug 2026
Concept: Retardation is negative acceleration. Acceleration $a$ is given by $a = \frac{dv}{dt} = v \frac{dv}{dx}$.
By differentiating time with respect to distance $\frac{dt}{dx}$, we get the reciprocal of velocity: $v = \left(\frac{dt}{dx}\right)^{-1}$. Retardation is then equal to $-a$.
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)
By differentiating time with respect to distance $\frac{dt}{dx}$, we get the reciprocal of velocity: $v = \left(\frac{dt}{dx}\right)^{-1}$. Retardation is then equal to $-a$.
A.
$2\alpha v^3$
B.
$2\beta v^3$
C.
$2\alpha\beta v^3$
D.
$2\beta^2 v^3$
Q142
Revision
MCQ
15 Aug 2026
Concept: Displacement $x$ as a function of time $t$ for constant acceleration is given by $x \propto t^2$ or $x = K t^2$, where $K$ is a constant.
Acceleration is the second time derivative of displacement:
$a = \frac{d^2 x}{d t^2}$
If the second derivative is a constant (independent of time), the particle moves with uniform acceleration.
If displacement of a particle is directly proportional to the square of time. Then particle is moving with
Acceleration is the second time derivative of displacement:
$a = \frac{d^2 x}{d t^2}$
If the second derivative is a constant (independent of time), the particle moves with uniform acceleration.
A.
Uniform acceleration
B.
Variable acceleration
C.
Uniform velocity
D.
Variable acceleration but uniform velocity
Q143
Revision
MCQ
15 Aug 2026
Concept: Average acceleration is defined as the change in velocity divided by the total time taken:
$\vec{a}_{av} = \frac{\Delta\vec{v}}{\Delta t} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t}$
The magnitude of the change in velocity for two orthogonal vectors is calculated using:
$\Delta v = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos(90^\circ)} = \sqrt{v_1^2 + v_2^2}$
The direction of $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ is determined by adding vector $\vec{v}_2$ (North) and vector $-\vec{v}_1$ (West), which points North-West.
A particle is moving eastwards with velocity of 5 m/s. In 10 sec the velocity changes to 5 m/s northwards. The average acceleration in this time is from the image given below
$\vec{a}_{av} = \frac{\Delta\vec{v}}{\Delta t} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t}$
The magnitude of the change in velocity for two orthogonal vectors is calculated using:
$\Delta v = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos(90^\circ)} = \sqrt{v_1^2 + v_2^2}$
The direction of $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ is determined by adding vector $\vec{v}_2$ (North) and vector $-\vec{v}_1$ (West), which points North-West.
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
Q144
Revision
MCQ
15 Aug 2026
Concept: Position $x$ is obtained by integrating velocity with respect to time:
$x = \int v \, dt$
Instantaneous acceleration $a$ is the time derivative of velocity:
$a = \frac{dv}{dt}$
First solve for the time $t$ when position $x = 2\text{ m}$, and then calculate the acceleration at that instant.
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 2m from the origin?
$x = \int v \, dt$
Instantaneous acceleration $a$ is the time derivative of velocity:
$a = \frac{dv}{dt}$
First solve for the time $t$ when position $x = 2\text{ m}$, and then calculate the acceleration at that instant.
A.
$28\text{ m/s}^2$
B.
$22\text{ m/s}^2$
C.
$12\text{ m/s}^2$
D.
$10\text{ m/s}^2$
Q145
Revision
MCQ
15 Aug 2026
Concept: Acceleration is defined as the rate of change of velocity over time:
$\vec{a} = \frac{\Delta \vec{v}}{t} = \frac{\vec{v}_2 - \vec{v}_1}{t}$
When the direction of motion reverses, the final velocity vector is taken as negative relative to the initial velocity vector.
A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 sec on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is
$\vec{a} = \frac{\Delta \vec{v}}{t} = \frac{\vec{v}_2 - \vec{v}_1}{t}$
When the direction of motion reverses, the final velocity vector is taken as negative relative to the initial velocity vector.
A.
$3\text{ m/s}^2$
B.
$-3\text{ m/s}^2$
C.
$0.3\text{ m/s}^2$
D.
$-0.3\text{ m/s}^2$
Q146
Revision
MCQ
15 Aug 2026
Concept: Average velocity over time intervals is given by $v = \frac{\Delta x}{\Delta t}$.
If the velocity increases continuously over successive equal time intervals, the motion is accelerated. If the rate of change of velocity (acceleration) is constant, it is uniformly accelerated; otherwise, it is non-uniformly accelerated.
The position of a particle moving along the x-axis at certain times is given below from the image given below:If the velocity increases continuously over successive equal time intervals, the motion is accelerated. If the rate of change of velocity (acceleration) is constant, it is uniformly accelerated; otherwise, it is non-uniformly accelerated.
$t (s)$: 0, 1, 2, 3$x (m)$: -2, 0, 6, 16
Which of the following describes the motion correctly
A.
Uniform, accelerated
B.
Uniform, decelerated
C.
Non-uniform, accelerated
D.
There is not enough data for generalisation
Q147
Revision
MCQ
15 Aug 2026
Concept: Uniform motion means a body travels equal distances in equal intervals of time along a straight line, which implies a constant velocity.
In a distance-time ($s-t$) graph, the slope of the curve represents speed ($v = \frac{ds}{dt}$).
A straight line inclined to the time axis indicates a constant slope, which signifies uniform speed or uniform motion.
Which of the following graph represents uniform motion from the image given below
In a distance-time ($s-t$) graph, the slope of the curve represents speed ($v = \frac{ds}{dt}$).
A straight line inclined to the time axis indicates a constant slope, which signifies uniform speed or uniform motion.
A.
B.
C.
D.
Q148
Revision
MCQ
15 Aug 2026
Concept: The velocity of a particle from a displacement-time graph is given by the slope of the straight line inclined to the time axis:
$v = \tan\theta$
where $\theta$ is the angle made by the displacement-time line with the time axis.
The displacement-time graph for two particles $A$ and $B$ are straight lines inclined at angles of $30^\circ$ and $60^\circ$ with the time axis. The ratio of velocities of $v_A : v_B$ is
$v = \tan\theta$
where $\theta$ is the angle made by the displacement-time line with the time axis.
A.
$1 : 2$
B.
$1 : \sqrt{3}$
C.
$\sqrt{3} : 1$
D.
$1 : 3$
Q149
Revision
MCQ
15 Aug 2026
Concept: Velocity from a position-time graph is given by the slope of the graph measured relative to the time axis:
$v = \tan\theta$
where $\theta$ is the angle made by the straight line with the time axis. If the angle given in the diagram is measured relative to the displacement axis ($\theta_{\text{disp}}$), the angle with respect to the time axis is $\theta = 90^\circ - \theta_{\text{disp}}$.
From the image given below find out the velocity of a moving body
$v = \tan\theta$
where $\theta$ is the angle made by the straight line with the time axis. If the angle given in the diagram is measured relative to the displacement axis ($\theta_{\text{disp}}$), the angle with respect to the time axis is $\theta = 90^\circ - \theta_{\text{disp}}$.
A.
$\frac{1}{\sqrt{3}}\text{ m/s}$
B.
$3\text{ m/s}$
C.
$\sqrt{3}\text{ m/s}$
D.
$\frac{1}{3}\text{ m/s}$
Q150
Revision
MCQ
15 Aug 2026
Concept: Average velocity over a given time interval is defined as total displacement divided by total time elapsed:
$v_{av} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{x_f - x_i}{t_f - t_i}$
where $x_i$ is the initial position at time $t_i$ and $x_f$ is the final position at time $t_f$.
The diagram shows the displacement-time graph for a particle moving in a straight line from the image given below. The average velocity for the interval $t = 0, t = 5$ is
$v_{av} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{x_f - x_i}{t_f - t_i}$
where $x_i$ is the initial position at time $t_i$ and $x_f$ is the final position at time $t_f$.
A.
$0$
B.
$6\text{ ms}^{-1}$
C.
$-2\text{ ms}^{-1}$
D.
$2\text{ ms}^{-1}$