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Kinematics-1D

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All Exams Advanced (28) Error less (56) Revision (69) DPT (54)
Q1 Error less Distance & Displacement MCQ
25 Jul 2026
Concept: Displacement is a vector quantity defined as the shortest straight-line distance from the initial position to the final position. If East is taken along the positive x-axis and North along the positive y-axis, the displacement vector is given by $\vec{r} = x\hat{i} + y\hat{j}$. Its magnitude is calculated using the formula $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2}$.
A man goes 10 m towards North, then 20 m towards East. What is his displacement?
A.
22.5 m
B.
25 m
C.
25.5 m
D.
30 m
Q2 Error less Distance & Displacement MCQ
25 Jul 2026
Concept: Distance is the actual total path length covered during motion, which for a quarter circle of radius $r$ is one fourth of the circumference, given by $s = \frac{2\pi r}{4} = \frac{\pi r}{2}$.
Displacement is the straight-line distance from the initial position to the final position. In a quarter circular path, the initial and final position vectors form a right-angled triangle with the center, so the magnitude of displacement is $d = \sqrt{r^2 + r^2} = r\sqrt{2}$.
A body moves over one fourth of a circular arc in a circle of radius $r$. The magnitude of distance travelled and displacement will be respectively
A.
$\frac{\pi r}{2}, r\sqrt{2}$
B.
$\frac{\pi r}{4}, r$
C.
$\pi r, \frac{r}{\sqrt{2}}$
D.
$\pi r, r$
Q3 Error less Distance & Displacement MCQ
25 Jul 2026
Concept: When a wheel rolls forward through half a revolution without slipping:
1. The horizontal displacement of the center of the wheel (and thus the whole wheel) is equal to half of its circumference, which is $x = \pi R$.
2. The point initially in contact with the ground moves to the top of the wheel after half a revolution, causing a vertical displacement equal to the diameter of the wheel, $y = 2R$.
The net displacement vector connects the initial ground contact point to its final top position. By the Pythagorean theorem, the magnitude of total displacement is $d = \sqrt{x^2 + y^2} = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$.
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 $R$)
A.
$\frac{R}{\sqrt{\pi^2 + 4}}$
B.
$R\sqrt{\pi^2 + 4}$
C.
$2\pi R$
D.
$\pi R$
Q4 Error less Avg Speed and Velocity MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance travelled divided by the total time taken. When a journey is divided into segments, the time for each segment is calculated using the formula $t = \frac{\text{distance}}{\text{speed}}$, and the overall average speed is given by $v_{\text{av}} = \frac{d_{\text{total}}}{t_{\text{total}}}$.
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.
$\frac{5v_1 v_2}{3v_1 + 2v_2}$
Q5 Error less Distance & Displacement MCQ
25 Jul 2026
Concept: Displacement is defined as the shortest distance between the initial position and the final position of an object. If a body starts its motion from a point and returns to the exact same point, its net displacement is zero, which makes its average velocity zero since $\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}$. However, the distance covered along the path is non-zero, meaning the speed increases.
A car accelerated from initial position and then returned at initial point, then
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
Q6 Error less Avg Speed and Velocity MCQ
25 Jul 2026
Concept: Average speed is calculated as the total distance covered divided by the total time elapsed: $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$. To find the distance travelled in a specific time interval, determine the distance covered in each leg of the journey using $d = v \times t$.
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
A.
5 km/h
B.
$\frac{25}{4} \text{ km/h}$
C.
$\frac{30}{4} \text{ km/h}$
D.
$\frac{45}{8} \text{ km/h}$
Q7 Error less x-t Graph MCQ
25 Jul 2026
Concept: Average velocity over a time interval $\Delta t$ is defined as $v_{\text{av}} = \frac{\Delta x}{\Delta t}$. If the velocity increases by equal amounts in equal time intervals, the acceleration is constant, which signifies uniform accelerated motion.
The position of a particle moving along the x-axis at certain times is given below:
image.png 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
Q8 Error less x-t Graph MCQ
25 Jul 2026
Concept: In uniform motion, an object covers equal distances in equal intervals of time, meaning its speed (or velocity) remains constant. Since speed is given by the slope of the distance-time (or displacement-time) graph ($v = \frac{ds}{dt}$), uniform motion is represented by a straight line with a constant positive slope in a distance-time graph.
Which of the following graphs represents uniform motion?
A.
image.png
B.
image.png
C.
image.png
D.
image.png
Q9 Error less x-t Graph MCQ
25 Jul 2026
Concept: In a displacement-time ($s-t$) graph, the slope of the line represents the velocity of the particle. The slope of a line inclined at an angle $\theta$ with the time axis is given by $\tan\theta$. Therefore, the velocity $v$ is equal to $\tan\theta$.
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
A.
$1 : 2$
B.
$1 : \sqrt{3}$
C.
$\sqrt{3} : 1$
D.
$1 : 3$
Q10 Error less x-t Graph MCQ
25 Jul 2026
Concept: In kinematics, velocity is the rate of change of displacement with respect to time, $v = \frac{dx}{dt}$, which is represented by the slope of the displacement-time graph with respect to the time axis. Therefore, $v = \tan\theta$, where $\theta$ is the angle made by the line with the time axis.
From the following displacement-time graph, find out the velocity of a moving body:
(A graph is given with time on the vertical axis and displacement on the horizontal axis, showing a straight line making an angle of $30^\circ$ with the displacement axis). image.png
A.
$\frac{1}{\sqrt{3}} \text{ m/s}$
B.
$3 \text{ m/s}$
C.
$\sqrt{3} \text{ m/s}$
D.
$\frac{1}{3} \text{ m/s}$
Q11 Error less x-t Graph MCQ
25 Jul 2026
Concept: Average velocity over a time interval is defined as the total displacement divided by the total time taken: $v_{\text{av}} = \frac{\Delta x}{\Delta t} = \frac{x(t_f) - x(t_i)}{t_f - t_i}$, where $x(t_f)$ is the final position at time $t_f$ and $x(t_i)$ is the initial position at time $t_i$.
The diagram shows the displacement-time graph for a particle moving in a straight line. The average velocity for the interval $t = 0$ to $t = 5$ is image.png
A.
$0 \text{ ms}^{-1}$
B.
$6 \text{ ms}^{-1}$
C.
$-2 \text{ ms}^{-1}$
D.
$2 \text{ ms}^{-1}$
Q12 Error less x-t Graph MCQ
25 Jul 2026
Concept: In a displacement-time graph, the speed of a moving body is represented by the magnitude of the slope of the graph, given by $v = \left\vert{} \frac{\Delta y}{\Delta x} \right\vert{} = \left\vert{} \frac{\text{Change in displacement}}{\text{Change in time}} \right\vert{}$.
Figure shows the displacement-time graph of a body. What is the ratio of the speed in the first second and that in the next two seconds? image.png
A.
$1 : 2$
B.
$1 : 3$
C.
$3 : 1$
D.
$2 : 1$
Q13 Error less v-t Graph MCQ
25 Jul 2026
Concept: When air resistance is not ignored, the air drag force always opposes the direction of motion. During upward motion, both gravity and air resistance act downwards, giving a higher magnitude of acceleration $(g + a)$. During downward motion, gravity acts downwards while air resistance acts upwards, resulting in a lower magnitude of acceleration $(g - a)$.
A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored?
A.
image.png
B.
image.png
C.
image.png
D.
image.png
Q14 Error less v-t Graph MCQ
25 Jul 2026
Concept: In a speed-time graph, the acceleration at any segment is given by the slope of the line representing that segment, $a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$. The maximum acceleration corresponds to the segment with the steepest positive slope.
A train moves from one station to another in 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is image.png
A.
$100 \text{ km h}^{-2}$
B.
$160 \text{ km h}^{-2}$
C.
$140 \text{ km h}^{-2}$
D.
$120 \text{ km h}^{-2}$
Q15 Error less v-t Graph MCQ
25 Jul 2026
Concept: The slope of a displacement-time ($s-t$) graph represents the instantaneous velocity ($v = \frac{ds}{dt}$). For a downward-opening parabolic $s-t$ graph, the slope decreases continuously at a constant rate, starting from a positive value, reaching zero at the peak, and then becoming increasingly negative. This indicates a uniform negative acceleration, represented by a straight line with a negative slope on a velocity-time graph.
The graph of displacement $v/s$ time shows a downward-opening parabolic curve starting from the origin. Its corresponding velocity-time graph will be: image.png
A.
image.png
B.
image.png
C.
image.png
D.
image.png
Q16 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Velocity is the time derivative of displacement, $v = \frac{dS}{dt}$. Acceleration is the time derivative of velocity, $a = \frac{dv}{dt}$. To find the velocity when acceleration is zero, set $a = 0$ to solve for time $t$, and substitute that $t$ into the velocity equation.
A particle moves along a straight line such that its displacement at any time $t$ is given by $S = t^3 - 6t^2 + 3t + 4$ metres. The velocity when the acceleration is zero is
A.
$3 \text{ ms}^{-1}$
B.
$-12 \text{ ms}^{-1}$
C.
$42 \text{ ms}^{-1}$
D.
$-9 \text{ ms}^{-1}$
Q17 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Velocity is the rate of change of displacement with respect to time, $v = \frac{dx}{dt}$. Acceleration is the rate of change of velocity with respect to time, $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$.
A body is moving according to the equation $x = at + bt^2 - ct^3$ where $x =$ displacement and $a, b$ and $c$ are constants. The acceleration of the body is
A.
$2b - 6ct$
B.
$a + 2bt$
C.
$2b + 6ct$
D.
$3b - 6ct^2$
Q18 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Acceleration is the second derivative of displacement with respect to time, $a = \frac{d^2x}{dt^2}$.
The displacement is given by $x = 2t^2 + t + 5$, the acceleration at $t = 2\text{ s}$ is
A.
$4\text{ m/s}^2$
B.
$8\text{ m/s}^2$
C.
$10\text{ m/s}^2$
D.
$15\text{ m/s}^2$
Q19 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Acceleration is given by $a = \frac{dv}{dt}$. If $a$ depends on time $t$, the acceleration is non-uniform.
The velocity of a body depends on time according to the equation $v = 20 + 0.1t^2$. The body is undergoing
A.
Uniform acceleration
B.
Uniform retardation
C.
Non-uniform acceleration
D.
Zero acceleration
Q20 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: If displacement $x \propto t^3$, then $x = kt^3$ for some constant $k$. Acceleration is obtained by differentiating $x$ twice with respect to time $t$.
The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is
A.
Increasing with time
B.
Decreasing with time
C.
Constant but not zero
D.
Zero
Q21 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Velocity and acceleration are independent kinematic quantities at any given instant. A body can have zero instantaneous velocity while simultaneously having a non-zero acceleration (for example, a ball at the highest point of its vertical projectile motion).
The correct statement from the following is
A.
A body having zero velocity will not necessarily have zero acceleration
B.
A body having zero velocity will necessarily have zero acceleration
C.
A body having uniform speed can have only uniform acceleration
D.
A body having non-uniform velocity will have zero acceleration
Q22 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: To find displacement when acceleration is zero, differentiate displacement twice to find acceleration, set $a = 0$ to find time $t$, and substitute $t$ back into the displacement equation.
A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^3 - 3t^2 + 2\text{ meter}$. The displacement when the acceleration becomes zero is
A.
$0\text{ meter}$
B.
$2\text{ meter}$
C.
$3\text{ meter}$
D.
$-2\text{ meter}$
Q23 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Retardation means acceleration acts in the direction opposite to velocity (and direction of motion/displacement).
What is the angle between instantaneous displacement and acceleration during the retarded motion
A.
Zero
B.
$\frac{\pi}{4}$
C.
$\frac{\pi}{2}$
D.
$\pi$
Q24 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Acceleration is the derivative of velocity with respect to time, $A = \frac{dv}{dt}$, and velocity is the derivative of displacement with respect to time, $v = \frac{dx}{dt}$. To find displacement from acceleration, integrate the acceleration function twice with respect to time, applying initial conditions $v(0) = 0$ and $x(0) = 0$.
The acceleration of a particle starting from rest, varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be
A.
$-\frac{1}{2}(a\omega^2 \sin\omega t)t^2$
B.
$a\omega \sin\omega t$
C.
$a\omega \cos\omega t$
D.
$a \sin\omega t$
Q25 Error less Acceleration & Calculus MCQ
28 Jul 2026
Concept: Average acceleration over a time interval from $t_1$ to $t_2$ is defined as the change in velocity divided by the total time taken: $a_{\text{avg}} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$.
If the velocity of a particle is $(10 + 2t^2) \text{ m/s}$, then the average acceleration of the particle between $2\text{ s}$ and $5\text{ s}$ is
A.
$2 \text{ m/s}^2$
B.
$4 \text{ m/s}^2$
C.
$12 \text{ m/s}^2$
D.
$14 \text{ m/s}^2$
Q26 Error less x-t Graph MCQ
29 Jul 2026
Concept: According to Newton's first law of motion, no net force acts on a body when its acceleration is zero ($a = 0$). In a displacement-time ($x-t$) graph, the slope represents velocity ($v = \frac{dx}{dt}$). Zero acceleration corresponds to a constant velocity, which is represented by a straight line with a constant slope in the displacement-time graph.
The displacement versus time graph for a body moving in a straight line is shown from the image given below. Which of the following regions represents the motion when no force is acting on the body? image.png
A.
$ab$
B.
$bc$
C.
$cd$
D.
$de$
Q27 Error less x-t Graph MCQ
29 Jul 2026
Concept: When an object undergoes uniform deceleration, its acceleration $a$ is negative and constant. The equation for displacement is $x = ut + \frac{1}{2}at^2$. Since velocity $v = \frac{dx}{dt}$ represents the slope of the displacement-time graph, uniform deceleration implies that the slope of the $x-t$ graph must decrease continuously over time.
A car decelerates at a constant rate during a period commencing at $t = 0$. Which of the displacement time graphs represents the displacement of the car from the image given below?
A.
(a) image.png
B.
(b) image.png
C.
(c) image.png
D.
(d) image.png
Q28 Error less x-t Graph MCQ
29 Jul 2026
Concept: Distance covered by a moving body can never decrease with time; it must either increase or remain constant if the body is at rest ($d \ge 0$, $\frac{dd}{dt} \ge 0$). Additionally, time always flows forward, so distance cannot have multiple values for a single instant of time. Therefore, any graph where distance decreases as time increases cannot represent a valid distance-time graph.
Which of the following can not be the distance time graph from the image given below?
A.
(a) image.png
B.
(b) image.png
C.
(c) image.png
D.
(d) image.png
Q29 Error less x-t Graph MCQ
29 Jul 2026
Concept: Time is an independent variable that always flows forward monotonically and cannot go backwards. Furthermore, a single particle cannot exist at two different positions at the exact same instant of time. Therefore, any displacement-time graph that shows time traveling backward or multiple displacement values for a single instant of time is physically impossible.
Which of the following displacement time graphs is not possible from the image given below?
A.
Graph (a) image.png
B.
Graph (b) image.png
C.
Graph (c) image.png
D.
Graph (d) image.png
Q30 Error less x-t Graph MCQ
29 Jul 2026
Concept: In a displacement-time ($x-t$) graph, the slope represents the velocity ($v = \frac{dx}{dt}$), and the curvature represents the acceleration ($a = \frac{d^2x}{dt^2}$). If the graph is concave downward (slope decreasing), the acceleration is negative ($-$). If the graph is a straight line (slope constant), the velocity is constant and acceleration is zero ($0$). If the graph is concave upward (slope increasing), the acceleration is positive ($+$).
The graph between the displacement $x$ and time $t$ for a particle moving in a straight line is shown from the image gven below. During the intervals $OA$, $AB$, $BC$, and $CD$, what is the sign of the acceleration of the particle? image.png
A.
$+$, $0$, $+$, $x$
B.
$-$, $0$, $+$, $0$
C.
$+$, $0$, $-$, $x$
D.
$-$, $0$, $-$, $0$
Q31 Error less x-t Graph MCQ
29 Jul 2026
Concept: In a displacement-time ($x-t$) graph, the slope represents the velocity of the body ($v = \frac{dx}{dt}$). A straight inclined line indicates motion with constant speed, while a horizontal line parallel to the time axis indicates that displacement is constant, meaning the body is at rest ($v = 0$).
The $x-t$ graph represents from the image gven below image.png
A.
Constant velocity
B.
Velocity of the body continuously changing
C.
Instantaneous velocity
D.
The body travels with constant speed upto time $t_1$ and then stops
Q32 Error less v-t Graph MCQ
29 Jul 2026
Concept: When an object moves with uniform acceleration $a$, the third equation of motion relating initial velocity $u$, final velocity $v$, acceleration $a$, and displacement $s$ is $v^2 = u^2 + 2as$. Assuming initial velocity $u = 0$, the equation becomes $v^2 = 2as$ or $s = \frac{v^2}{2a}$. This represents a parabolic equation where $s \propto v^2$, opening towards the $s$-axis.
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement $s$ -velocity $v$ graph of this object is from the image given below
A.
A parabola opening towards the $s$-axis image.png
B.
A curve curving towards the $v$-axis image.png
C.
A parabola opening upwards along $s$-axis starting after a constant offset image.png
D.
A straight line passing through origin image.png
Q33 Error less v-t Graph MCQ
29 Jul 2026
Concept: The total distance travelled by a particle in a given time interval is equal to the total area under the velocity-time graph for that duration. For any interval, the area of a trapezoid formed under the graph is given by $\text{Area} = \frac{1}{2} \times (v_1 + v_2) \times \Delta t$, where $v_1$ and $v_2$ are the initial and final velocities during time interval $\Delta t$.
The variation of velocity of a particle with time moving along a straight line is illustrated in the graph from the image given below. The distance travelled by the particle in four seconds is image.png
A.
60 m
B.
55 m
C.
25 m
D.
30 m
Q34 Error less v-t Graph MCQ
29 Jul 2026
Concept: Let $t_1$ be the time during acceleration and $t_2$ be the time during deceleration, such that $t = t_1 + t_2$. For motion starting from rest with uniform acceleration $\alpha$, the maximum velocity is $v_{\text{max}} = \alpha t_1$. For deceleration back to rest at rate $\beta$, the maximum velocity is also $v_{\text{max}} = \beta t_2$. Substituting $t_1 = \frac{v_{\text{max}}}{\alpha}$ and $t_2 = \frac{v_{\text{max}}}{\beta}$ into $t = t_1 + t_2$ gives $t = v_{\text{max}} \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$. Rearranging for $v_{\text{max}}$ yields $v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta}$.
A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed in $t$, then the maximum velocity acquired by the car is
A.
$(\frac{\alpha^{2}+\beta^{2}}{\alpha\beta})t$
B.
$(\frac{\alpha^{2}-\beta^{2}}{\alpha\beta})t$
C.
$\frac{(\alpha+\beta)t}{\alpha\beta}$
D.
$\frac{\alpha\beta t}{\alpha+\beta}$
Q35 Error less v-t Graph MCQ
29 Jul 2026
Concept: The maximum height reached by the rocket corresponds to the total area under the positive region of the velocity-time graph up to the point where the velocity becomes zero. For a triangle on a velocity-time graph with base $b$ and height $h$, the maximum height reached is given by the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
A rocket is projected vertically upwards, whose velocity-time graph is shown in fig. The maximum height reached by the rocket is from the image given below image.png
A.
1 km
B.
10 km
C.
20 km
D.
60 km
Q36 Error less v-t Graph MCQ
29 Jul 2026
Concept: Mean or average velocity is defined as the total displacement divided by the total time taken. The formula is $\text{Mean velocity} = \frac{\text{Total displacement}}{\text{Total time}}$.
In the above problem the mean velocity of rocket in reaching the maximum height will be from the image given below A rocket is projected vertically upwards, whose velocity-time graph is shown in fig. image.png
A.
100 m/s
B.
50 m/s
C.
500 m/s
D.
25/3 m/s
Q37 Error less v-t Graph MCQ
29 Jul 2026
Concept: Acceleration is defined as the rate of change of velocity with respect to time. On a velocity-time graph, the acceleration corresponds to the slope of the velocity-time curve during the period of speed increase. The formula is $a = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}$.
In the above problem the acceleration of rocket will be from the image given below A rocket is projected vertically upwards, whose velocity-time graph is shown in fig. image.png
A.
$50 \text{ m/s}^2$
B.
$100 \text{ m/s}^2$
C.
$500 \text{ m/s}^2$
D.
$250 \text{ m/s}^2$
Q38 Error less v-t Graph MCQ
29 Jul 2026
Concept: The total height (displacement) reached by the lift is equal to the total area under its velocity-time graph. The graph forms a trapezium with parallel sides representing the time duration at maximum speed $b = 10 - 2 = 8 \text{ s}$ and total time $a = 12 \text{ s}$, with maximum height (velocity) $h = 3.6 \text{ m/s}$. The area of a trapezium is given by $\text{Area} = \frac{1}{2} \times (a + b) \times h$.
A lift is going up. The variation in the speed of the lift is as given in the graph. What is height to which the lift takes the passenger from the image given below image.png
A.
3.6 m
B.
28.8 m
C.
36.0 m
D.
Cannot be calculated from the above graph
Q39 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 image.png
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
Q40 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 image.png
A.
0
B.
$2.5 \text{ ms}^{-1}$
C.
$5 \text{ ms}^{-1}$
D.
$2 \text{ ms}^{-1}$
Q41 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.
image.png
B.
image.png
C.
image.png
D.
image.png
Q42 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
Q43 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 image.png
A.
image.png
B.
image.png
C.
image.png
D.
image.png
Q44 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 image.png
A.
image.png
B.
image.png
C.
image.png
D.
image.png
Q45 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.
image.png
B.
image.png
C.
image.png
D.
image.png
Q46 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 image.png
A.
200
B.
250
C.
300
D.
400
Q47 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? image.png
A.
1/2
B.
1/4
C.
1/3
D.
2/3
Q48 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 image.png
A.
8 m, 16 m
B.
16 m, 8 m
C.
16 m, 16 m
D.
8 m, 8 m
Q49 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?
A.
image.png
B.
image.png
C.
image.png
D.
image.png
Q50 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:
A.
image.png
B.
image.png
C.
image.png
D.
image.png