iCON Education - Motion in 1D

Q101 Objective Physics Vol-1 Motion under Gravity MCQ

iCON Education HYD, 79930 92826, 73309 72826 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

Q102 Objective Physics Vol-1 Motion under Gravity MCQ

iCON Education HYD, 79930 92826, 73309 72826 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

Q103 Objective Physics Vol-1 2. Uniform Accelerated MCQ

iCON Education HYD, 79930 92826, 73309 72826 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.

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

Q104 Objective Physics Vol-1 2. Uniform Accelerated MCQ

iCON Education HYD, 79930 92826, 73309 72826 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

Q105 Objective Physics Vol-1 2. Uniform Accelerated MCQ

iCON Education HYD, 79930 92826, 73309 72826 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

Q106 Objective Physics Vol-1 2. Uniform Accelerated MCQ

iCON Education HYD, 79930 92826, 73309 72826 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.

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

Q107 Objective Physics Vol-1 2. Uniform Accelerated MCQ

iCON Education HYD, 79930 92826, 73309 72826 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 $\text{ms}^{-1}$

Q108 Objective Physics Vol-1 2. Uniform Accelerated MCQ

iCON Education HYD, 79930 92826, 73309 72826 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 $\text{ms}^{-1}$

Q109 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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?

A.

Zero

B.

Negative

C.

Positive

D.

Cannot be determined

Q110 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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?

A.

Positive constant

B.

Negative constant

C.

Zero

D.

Variable

Q111 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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:

A.

Positive

B.

Negative

C.

Zero

D.

Infinite

Q112 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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?

A.

Velocity is constant

B.

Velocity is decreasing

C.

Velocity is increasing with positive acceleration

D.

Velocity is increasing with negative acceleration

Q113 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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

Q114 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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 $t = 3 \text{ s}$).

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

Q115 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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 $\frac{\alpha \beta t^2}{\alpha + \beta}$

Q116 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 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?

A.

$60 \mathrm{m}$

B.

$70 \mathrm{m}$

C.

$80 \mathrm{m}$

D.

$100 \mathrm{m}$

Q117 Errorless 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The slope of a displacement-time graph represents the velocity of the particle.
$\text{Velocity } v = \frac{dx}{dt}$
If the slope decreases over time, the velocity is decreasing, which indicates retarded motion until the slope becomes zero and the particle stops.

The displacement of a particle as a function of time is shown in the image given below. The figure shows that

A.

The particle starts with certain velocity but the motion is retarded and finally the particle stops

B.

The velocity of the particle is constant throughout

C.

The acceleration of the particle is constant throughout

D.

The particle starts from rest and accelerates uniformly

Q118 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The motion of the rocket can be divided into three phases: upward acceleration with fuel, upward deceleration under gravity after fuel finishes, and downward acceleration under gravity. The total time is the sum of the time taken in each phase. The area under the velocity-time graph gives the maximum height, which is then used to find the time for the downward journey.

A rocket is fired vertically upwards with a net acceleration of $4 \, \text{ms}^{-2}$ and initial velocity zero. After 5 s, its fuel is finished and it decelerates with $g$. At the highest point, its velocity becomes zero. Then, it accelerates downwards with acceleration $g$ and returns back to the ground. What is the total time taken for the complete journey? (Take, $g = 10 \, \text{ms}^{-2}$)

A.

$7 + \sqrt{14} \, \text{s}$

B.

$5 + \sqrt{14} \, \text{s}$

C.

$10 + \sqrt{14} \, \text{s}$

D.

$12 + \sqrt{14} \, \text{s}$

Correct Answer: A

Explanation:

Step 1: For the first phase (with fuel), initial velocity $u = 0$, acceleration $a = 4 \, \text{ms}^{-2}$, and time $t_{OA} = 5 \, \text{s}$.

Step 2: The velocity attained after 5 s is $v_A = u + at_{OA} = 0 + (4)(5) = 20 \, \text{ms}^{-1}$.
Step 3: The distance covered in this phase is $s_{OA} = \frac{1}{2} a t_{OA}^2 = \frac{1}{2} (4)(5)^2 = 50 \, \text{m}$.
Step 4: For the second phase (fuel finished, moving up), initial velocity $u = 20 \, \text{ms}^{-1}$, final velocity $v_B = 0$, and acceleration $a = -g = -10 \, \text{ms}^{-2}$.
Step 5: The time taken to reach the highest point is $t_{AB} = \frac{v_B - v_A}{-g} = \frac{0 - 20}{-10} = 2 \, \text{s}$.
Step 6: The total time to reach the maximum height is $t_{OAB} = t_{OA} + t_{AB} = 5 + 2 = 7 \, \text{s}$.
Step 7: The maximum height reached is the area under the velocity-time graph up to the highest point: $H = s_{OAB} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (7)(20) = 70 \, \text{m}$.
Step 8: For the third phase (falling down from the highest point), initial velocity $u = 0$, distance $s_{BC} = 70 \, \text{m}$, and acceleration $a = g = 10 \, \text{ms}^{-2}$.
Step 9: Using the equation $s = ut + \frac{1}{2}at^2$, we get $70 = 0 + \frac{1}{2}(10)t_{BC}^2$, which simplifies to $5t_{BC}^2 = 70$.
Step 10: Solving for $t_{BC}$, we get $t_{BC}^2 = 14$, so $t_{BC} = \sqrt{14} \, \text{s}$.
Step 11: The total time for the complete journey is $t_{total} = t_{OAB} + t_{BC} = 7 + \sqrt{14} \, \text{s}$.

Q119 Objective Physics Vol-1 6. Graphical Analysis (a-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The area under the acceleration-time (a-t) graph gives the change in velocity of the particle over the given time interval. By adding this change in velocity to the initial velocity, we can find the final velocity.

The acceleration-time graph of a particle moving in a straight line is a triangle with a base of 4 s and a height of $4 \, \text{ms}^{-2}$. If the velocity of the particle at time $t = 0$ is $2 \, \text{ms}^{-1}$, what is the velocity of the particle at the end of the fourth second?

A.

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

B.

$8 \, \text{ms}^{-1}$

C.

$10 \, \text{ms}^{-1}$

D.

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

Q120 Objective Physics Vol-1 6. Graphical Analysis (a-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The velocity of a particle at any instant can be found by calculating the area under the acceleration-time graph, or by applying the kinematic equation $v = u + at$ for each time interval where acceleration is constant.

The acceleration versus time graph of a particle moving along a straight line is given such that $a = +2 \text{ m/s}^2$ from $t = 0$ to $t = 2 \text{ s}$, $a = 0$ from $t = 2$ to $t = 4 \text{ s}$, and $a = -4 \text{ m/s}^2$ from $t = 4$ to $t = 6 \text{ s}$. Assuming the initial velocity at $t = 0$ is $0 \text{ m/s}$, what is the velocity of the particle at the end of $6 \text{ s}$?

A.

$0 \text{ m/s}$

B.

$-4 \text{ m/s}$

C.

$4 \text{ m/s}$

D.

$-8 \text{ m/s}$

Q121 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: Uniform motion is defined as motion with a constant velocity. In a displacement-time ($s-t$) graph, uniform motion is represented by a straight line because the slope (which represents velocity, $v = \frac{ds}{dt}$) remains constant.

Which of the following graph represents the uniform motion?

A.

B.

C.

D.

None of these

Q122 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: Instantaneous velocity is the slope of the distance-time (s-t) graph. The maximum instantaneous velocity corresponds to the point where the slope of the tangent to the curve is maximum (steepest).

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

A.

A

B.

B

C.

C

D.

D

Q123 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The slope of a displacement-time ($s-t$) graph represents the instantaneous velocity of the moving body. The slope is equal to the tangent of the angle made by the graph with the time axis.

From the displacement-time graph, find out the velocity of a moving body. (Assume the graph is a straight line making an angle of $30^\circ$ with the displacement axis).

A.

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

B.

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

C.

$\sqrt{3} \mathrm{ms}^{-1}$

D.

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

Q124 Objective Physics Vol-1 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The slope of a distance-time graph represents the instantaneous velocity of the particle. Average acceleration is defined as the rate of change of velocity over a given time interval.

The distance-time graph of a particle at time t makes an angle $45^\circ$ with the time axis. After one second, it makes an angle $60^\circ$ with the time axis. What is the average acceleration of the particle?

A.

$\sqrt{3}$

B.

$\sqrt{3} + 1$

C.

$\sqrt{3} - 1$

D.

$1$

Q125 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The acceleration of an object is given by the slope of its velocity-time (v-t) graph. The maximum acceleration corresponds to the maximum change in velocity in the minimum time interval, which is represented by the steepest slope on the graph.

The velocity-time graph of a moving object shows its maximum slope occurring in the time interval from $t = 30 \text{ s}$ to $t = 40 \text{ s}$, where the velocity changes from $20 \text{ cm/s}$ to $80 \text{ cm/s}$. What is the maximum acceleration of the object?

A.

$1 \text{ cm s}^{-2}$

B.

$2 \text{ cm s}^{-2}$

C.

$3 \text{ cm s}^{-2}$

D.

$6 \text{ cm s}^{-2}$

Q126 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The distance travelled by a particle is equal to the area under its velocity-time (v-t) graph. By calculating the area of the geometric shapes formed under the graph for each time interval and summing them up, we can find the total distance.

The variation of velocity of a particle with time moving along a straight line is illustrated in the adjoining figure. The distance travelled by the particle in 4 s is

A.

60 m

B.

55 m

C.

25 m

D.

30 m

Q127 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The distance travelled or height reached by an object is equal to the area under its velocity-time graph. For a trapezoidal velocity-time graph, the area is calculated using the formula for the area of a trapezium.

A lift is going up. The variation in the speed of the lift is as given in the graph, which forms a trapezium with parallel sides of 12 s and 8 s, and a maximum speed of 3.6 m/s. What is the height to which the lift takes the passengers?

A.

3.6 m

B.

28.8 m

C.

36.0 m

D.

Cannot be calculated from the above graph

Q128 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: Displacement is the net area under the velocity-time graph (area above the time axis minus area below), while distance is the total area under the velocity-time graph (sum of the absolute values of all areas).

The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 s are respectively

A.

$8 \mathrm{m}, 16 \mathrm{m}$

B.

$16 \mathrm{m}, 32 \mathrm{m}$

C.

$16 \mathrm{m}, 16 \mathrm{m}$

D.

$8 \mathrm{m}, 18 \mathrm{m}$

Q129 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The velocity of a particle is the rate of change of its displacement with respect to time, given by $v = \frac{dx}{dt}$. By differentiating the displacement-time equation, we can find the velocity-time equation and determine the nature of the $v-t$ graph.

The $x - t$ equation is given as $x = 2t + 1$. The corresponding $v - t$ graph is

A.

a straight line passing through origin

B.

a straight line not passing through origin

C.

a parabola

D.

None of the above

Q130 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: For a particle released from rest, the initial velocity is zero. Under free fall, it experiences a constant acceleration due to gravity. The velocity-time relation is given by $v = gt$, which represents a straight line passing through the origin with a constant slope.

Which of the following graphs correctly represents the velocity-time relationship for a particle released from rest to fall freely under gravity?

A.

B.

C.

D.

Q131 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: When a particle is projected vertically upwards, it experiences a constant downward acceleration due to gravity ($a = -g$). The velocity-time relation is given by $v = u - gt$, which is a linear equation. The velocity is initially positive, decreases linearly to zero at the highest point, and then becomes negative as it falls back down, increasing in magnitude linearly.

A particle is projected vertically upwards and returns to the ground in time $T$. Which of the following graphs correctly represents the variation of velocity ($v$) against time ($t$)?

A.

B.

C.

D.

Q132 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The acceleration of a particle is equal to the slope of its velocity-time (v-t) graph. The slope is calculated as the change in velocity divided by the change in time.

The velocity-time graph for a particle is a straight line starting from $15 \text{ m/s}$ at $t = 0$ and reaching $0 \text{ m/s}$ at $t = 3 \text{ s}$. What is the acceleration of the particle?

A.

$225 \text{ ms}^{-2}$

B.

$5 \text{ ms}^{-2}$

C.

$-5 \text{ ms}^{-2}$

D.

$-3 \text{ ms}^{-2}$

Q133 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The average velocity of an object over a time interval is given by the total displacement divided by the total time taken. The displacement is equal to the net area under the velocity-time (v-t) graph (area above the time axis minus area below the time axis).

The v-t plot of a moving object is shown in the figure. The average velocity of the object during the first 10 s is

A.

zero

B.

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

C.

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

D.

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

Q134 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: In one-dimensional motion, distance and speed are scalar quantities that cannot be negative or decrease with time. Furthermore, at any given instant of time, a particle can only have one unique value of position and one unique value of velocity. Therefore, graphs violating these conditions are physically impossible.

Which of the following graphs cannot possibly represent one dimensional motion of a particle?

A.

I and II

B.

II and III

C.

II and IV

D.

All of these

Q135 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: The relationship between acceleration, velocity, and displacement is given by $a = v \frac{dv}{ds}$. When velocity decreases linearly with displacement, the $v$-$s$ graph is a straight line with a constant negative slope, which represents the derivative $\frac{dv}{ds}$.

If the velocity $v$ of a particle moving along a straight line decreases linearly with its displacement $s$ from $20 \text{ ms}^{-1}$ to a value approaching zero at $s = 30 \text{ m}$, then what is the acceleration of the particle at $v = 10 \text{ ms}^{-1}$?

A.

$\frac{2}{3} \text{ ms}^{-2}$

B.

$-\frac{2}{3} \text{ ms}^{-2}$

C.

$\frac{20}{3} \text{ ms}^{-2}$

D.

$-\frac{20}{3} \text{ ms}^{-2}$

Q136 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: For uniformly accelerated motion, the kinematic equation $v^2 = u^2 + 2as$ shows that a $v^2$ versus $s$ graph is a straight line with intercept $u^2$ on the $v^2$ axis and slope $2a$. A non-zero intercept implies non-zero initial velocity.

The $v^2$ versus $s$ graph of a particle moving in a straight line is a straight line with a non-zero intercept on the $v^2$ axis. From the graph, some conclusions are drawn. State which amongst the following statement(s) is wrong?

A.

The given graph shows a uniformly accelerated motion.

B.

Initial velocity of particle is zero.

C.

Corresponding s-t graph will be a parabola.

D.

None of the above

Q137 Objective Physics Vol-1 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: Kinematic equation of motion relating velocity, acceleration, and displacement, and the interpretation of the slope of a $v^2$ versus $s$ graph.

A graph between the square of the velocity of a particle and the distance s moved by the particle is shown in the figure below. The acceleration of the particle is

A.

$-8\mathrm{ms}^{-2}$

B.

$-4\mathrm{ms}^{-2}$

C.

$-16\mathrm{ms}^{-2}$

D.

None of these

Q138 Objective Physics Vol-1 6. Graphical Analysis (a-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 30 Jul 2026

Concept: Integration of an acceleration-time graph to determine the corresponding velocity-time graph, using the kinematic equation $v = u + at$.

A particle starts from rest at t = 0 and undergoes an acceleration a in ms $^{-2}$ with time t in second which is as shown. Which one of the following plot represents velocity v (in ms $^{-1}$ ) versus time t (in s)?

A.

B.

C.

D.

Q139 Allen 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Distance is the total path length travelled by a particle. In a velocity-time graph, the distance travelled in a given time interval is equal to the total area under the $v-t$ curve, taking all area magnitudes as positive (i.e., taking the absolute value of negative areas).
Formula:
$\text{Distance} = \int \vert{}v\vert{} dt = \text{Sum of magnitude of all areas under } v-t \text{ graph}$

From the image given below, the velocity-time graph of an object is shown. The distance during the interval 0 to $t_4$ is :-

A.

Area A + Area B + Area C + Area D + Area E

B.

Area A + Area C - Area B - Area D

C.

Area A + Area B + Area C + Area D

D.

Area A + Area C + Area E - Area B + Area D

Q140 Allen 6. Graphical Analysis (a-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Acceleration is defined as the rate of change of velocity with respect to time, which can be written as $a = \frac{dv}{dt}$ or $dv = a dt$.
The area under an acceleration-time ($a-t$) graph over a time interval represents the integration of acceleration with respect to time, which gives the total change in velocity during that time interval.
Formula:
$\text{Shaded Area} = \int a dt = \Delta v = v_{\text{final}} - v_{\text{initial}}$

From the image given below, the acceleration-time graph of a one dimensional motion is shown. Which of the following characteristics of the particle is represented by the shaded area?

A.

change in velocity

B.

change in position

C.

change in momentum

D.

velocity

Q141 Allen 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Velocity is defined as the rate of change of displacement with respect to time. On a displacement-time graph, velocity corresponds to the slope of the curve at any given point. When the velocity is zero, the slope of the tangent to the displacement-time curve is zero (horizontal tangent), which occurs at the local peak or crest of the graph.
Formula:
$v = \frac{ds}{dt} = \text{Slope of displacement-time graph}$

From the image given below, the displacement of a particle moving along x-axis is shown as a function of time. The velocity of the particle is zero at :-

A.

A

B.

B

C.

C

D.

D

Q142 Allen 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Instantaneous velocity is given by the rate of change of displacement with respect to time, which equals the slope of the tangent to the displacement-time graph at any point. A positive slope indicates positive velocity, a negative slope indicates negative velocity, and a horizontal tangent indicates zero velocity.
Formula:
$v = \frac{ds}{dt} = \text{Slope of the tangent on the displacement-time graph}$

From the image given below, the displacement-time graph of a moving particle is shown. The instantaneous velocity of the particle is positive at the point :-

A.

D

B.

F

C.

C

D.

E

Q143 Allen 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Average speed is defined as the total distance covered divided by the total time taken for the motion.
Formula: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$
Distance is calculated by taking the sum of the magnitudes of change in position during each segment of time.

A person walks along an east-west street and a graph of his displacement from home is shown from the image given below. His average speed for the whole time interval is

A.

0

B.

23 m/s

C.

8 m/s

D.

None of these

Q144 Allen 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Speed is defined as the magnitude of velocity, i.e., $\text{Speed} = \vert{}v\vert{}$. The speed is minimum when the absolute value of velocity $\vert{}v\vert{}$ reaches its lowest possible value, which is zero. On a velocity-time ($v-t$) graph, speed is zero whenever the velocity curve crosses or touches the time axis ($v = 0$).
Formula:
$\text{Speed} = \vert{}v\vert{}$
$\text{Minimum Speed} = 0 \text{ when } v = 0$

From the image given below, a particle is moving in a straight line $y=3x$. Its velocity time graph is shown in figure. Its speed is minimum at $t =$ .............

A.

2s

B.

4s

C.

6s

D.

8s

Q145 Allen 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 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 making an angle $\theta$ with the time axis (horizontal axis) is given by $\tan\theta$.
Formula:
$v = \text{Slope} = \tan\theta$
$\frac{v_A}{v_B} = \frac{\tan\theta_A}{\tan\theta_B}$

From the image given below, 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 velocity of particle A & B ($v_A : v_B$) is :-

A.

$1 : \sqrt{3}$

B.

$1 : 2$

C.

$\sqrt{3} : 1$

D.

$1 : 3$

Q146 Allen 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: In one-dimensional motion, a particle can have only one unique value of velocity at any given instant of time $t$. A graph that shows two or more values of velocity for a single time instant is physically impossible and cannot represent motion in one dimension.
Formula:
$\text{Unique velocity at any time } t \implies v(t) \text{ must be a single-valued function}$

From the image given below, which one of the following curves do not represent motion in one dimension :-

A.

B.

C.

D.

Q147 Allen 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: In a displacement-time ($s-t$) graph representing a realistic physical motion:
1. Time cannot flow backwards (time must always increase monotonically).
2. A body cannot exist in two or more different positions at the exact same instant of time.
3. The slope of the displacement-time graph represents velocity ($v = \frac{ds}{dt}$), which must remain finite. Infinite slope implies infinite velocity, which is physically impossible.
Formula:
$v = \frac{ds}{dt} = \text{Slope of } s-t \text{ graph}$

From the image given below, which of the following displacement-time graphs shows a realistic situation for a body in motion ?

A.

B.

C.

D.

Q148 Allen 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Uniformly accelerated motion is motion with a constant acceleration. Since acceleration is equal to the slope of the velocity-time ($v-t$) graph, constant acceleration is represented by a straight line with a constant slope.
Formula:
$a = \frac{dv}{dt} = \text{constant (slope of } v-t \text{ graph)}$

From the image given below, which of the following velocity-time graphs represent uniformly accelerated motion ?

A.

(1)

B.

(2)

C.

(3)

D.

(4)

Q149 Allen 4. Graphical Analysis (x-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: Average speed is defined as the total distance travelled by a particle divided by the total time taken. Distance is equal to the sum of the magnitudes of individual displacements in each motion segment.
Formula:
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$
$\text{Total distance} = \vert{}\Delta x_1\vert{} + \vert{}\Delta x_2\vert{}$

From the image given below, the figure shows the position time graph of a particle moving on a straight line path. What is the magnitude of average speed of the particle over 10 seconds ?

A.

2 m/s

B.

4 m/s

C.

6 m/s

D.

8 m/s

Q150 Allen 5. Graphical Analysis (v-t) MCQ

iCON Education HYD, 79930 92826, 73309 72826 01 Aug 2026

Concept: In a velocity-time graph:
1. Distance is the total area under the $v-t$ curve taking all area magnitudes as positive (absolute values).
2. Displacement is the net area under the $v-t$ curve taking area above the time axis as positive and area below the time axis as negative.
Formula:
$\text{Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 + \text{Area}_4$
$\text{Displacement} = (\text{Area}_1 + \text{Area}_2 + \text{Area}_3) - \text{Area}_4$
$\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area of Rectangle} = \text{length} \times \text{breadth}$

From the image given below, the velocity time graph of a body is shown. The distance travelled by the body and its displacement during 5 seconds in metres will be :-

A.

75, 75

B.

110, 40

C.

110, 110

D.

110, 70

Physics-I Chapters

Motion in 1D

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