NEET Kinematics MCQs & PYQs

Q: The given graph shows the variation of velocity with which one of the graph given below correctly represents the variation of acceleration with displacement :

\[ a= v \frac{dv}{ds} \] From the given $v-x$ graph, the equation of the straight line is: $$v = -\left(\frac{v_0}{x_0}\right)x + v_0$$ Acceleration $a$ as a function of position $x$ is defined as: $$a = v \frac{dv}{dx}$$ Find the velocity gradient $\frac{dv}{dx}$: $$\frac{dv}{dx} = -\frac{v_0}{x_0}$$ Substitute $v$ and $\frac{dv}{dx}$ into the acceleration equation: $$a = \left(-\frac{v_0}{x_0}x + v_0\right) \left(-\frac{v_0}{x_0}\right)$$ $$a = \left(\frac{v_0}{x_0}\right)^2 x - \frac{v_0^2}{x_0}$$ Conclusion: The $a-x$ relation is a straight line with a positive slope $m = \left(\frac{v_0}{x_0}\right)^2 > 0$ and a negative $y$-intercept $c = -\frac{v_0^2}{x_0}$. Therefore, the graph starts at a negative value on the acceleration axis ($a = -v_0^2 / x_0$ at $x = 0$) and increases linearly to zero at $x = x_0$ (represented by the first option).

Q: If the velocity of a particle is given by \[v=(180-16x)^{1/2} ms^{-1}\], then its acceleration will be

Given the velocity: v=180−16x m/s To find the acceleration, use the chain rule: a=dvdt=dvdx⋅dxdt=dvdx⋅v Differentiate vv v with respect to xx x: dvdx=−8180−16x\frac{dv}{dx} = \frac{-8}{\sqrt{180 - 16x}} Now, substitute v=180−16xv = \sqrt{180 - 16x} a=−8 m/s2 Thus, the acceleration is a=−8 m/s2a = -8 \, \text{m/s}^2

Q: Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1 : 8 : 1 and the maximum speed attained be 60 km h–¹, then what is the average speed over the whole journey?

Time Ratios: The time ratios are given as 1:8:11:8:1 1:8:1. So, if the total time is tt t, the train spends t10\frac{t}{10} 10t accelerating, 8t10\frac{8t}{10} 108t at constant velocity, and t10\frac{t}{10} 10t decelerating. Velocity-Time Graph: The graph forms a trapezium. The train accelerates linearly from 0 to 60 km/h, holds constant at 60 km/h, and then decelerates back to 0. Key points: The area under this graph represents the total distance traveled. The height (maximum velocity) = 60 km/h. The time intervals are in the ratio 1:8:11:8:1 1:8:1. Average Speed: The area of the trapezium is given by: Area=12×(ttotal)×(initial velocity+final velocity)\text{Area} = \frac{1}{2} \times (t_{\text{total}}) \times (\text{initial velocity} + \text{final velocity})Area=21×(ttotal)×(initial velocity+final velocity)For the constant velocity portion: Average speed=60×(1+8+1)10=54 km/h\text{Average speed} = \frac{60 \times (1 + 8 + 1)}{10} = 54 \, \text{km/h}Average speed=1060×(1+8+1)=54km/h Thus, the average speed is 54 km/h.

Q: A particle is thrown upwards from ground. It experience a constant air resistance force which can produce a retardation of 2 m/s². The ratio of time of ascent to the time of descent is : [g = 10 m/s²]

Effective accelerations: - Ascent: $ a_{\text{eff}} = g + a = 10 + 2 = 12 \, \text{m/s}^2 $ - Descent: $ a'_{\text{eff}} = g - a = 10 - 2 = 8 \, \text{m/s}^2 $ Ratio of times: \[ \frac{t_a}{t_d} = \sqrt{\frac{a'_{\text{eff}}}{a_{\text{eff}}}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}} \] The ratio of time of ascent to time of descent is √(2/3).

Question 1

The given graph shows the variation of velocity with which one of the graph given below correctly represents the variation of acceleration with displacement :

Answer

\[ a= v \frac{dv}{ds} \]

From the given

v-x

graph, the equation of the straight line is:

v = -\left(\frac{v_0}{x_0}\right)x + v_0

Acceleration

a

as a function of position

x

is defined as:

a = v \frac{dv}{dx}

Find the velocity gradient $\frac{dv}{dx}$ :

$\frac{dv}{dx} = -\frac{v_0}{x_0}$
Substitute $v$ and $\frac{dv}{dx}$ into the acceleration equation:

$a = \left(-\frac{v_0}{x_0}x + v_0\right) \left(-\frac{v_0}{x_0}\right)$

$a = \left(\frac{v_0}{x_0}\right)^2 x - \frac{v_0^2}{x_0}$

Conclusion:

The $a-x$ relation is a straight line with a positive slope $m = \left(\frac{v_0}{x_0}\right)^2 > 0$ and a negative $y$ -intercept $c = -\frac{v_0^2}{x_0}$ .
Therefore, the graph starts at a negative value on the acceleration axis ( $a = -v_0^2 / x_0$ at $x = 0$ ) and increases linearly to zero at $x = x_0$ (represented by the first option).

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Question 2

A particle is moving with a constant speed v on a circle of radius R, then find average acceleration of particle during half cycle is :

Answer

Change in velocity = 2V

Time taken = πR/V so,

Average acceleration = 2V/(πR/V)

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Question 3

Displacement (x) of a particle is related to time (t) as x = at + bt² – ct³, where a, b and c are constants of the motion. The velocity of the particle when its acceleration is zero is given by

Answer

Given:

\[
x = at + bt^2 - ct^3
\]

1. Velocity:

\[
v = \frac{dx}{dt} = a + 2bt - 3ct^2
\]

2.Acceleration:

\[
a = \frac{dv}{dt} = 2b - 6ct
\]

3. Set acceleration to zero:

\[
2b - 6ct = 0 ; t = \frac{b}{3c}
\]

4. Velocity at $t = \frac{b}{3c}$

\[
v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2
\]

\[
= a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c}
\]

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Question 4

A particle has an initial velocity \[3\hat{i}+4\hat{j}\] and an acceleration of \[0.4\hat{i}+0.3\hat{j}\] . Its speed after 10 s is

Answer

Given:

- Initial velocity: $\mathbf{u} = 3\hat{i} + 4\hat{j}$
- Acceleration: $\mathbf{a} = 0.4\hat{i} + 0.3\hat{j}$

Calculate final velocity after 10 s:

\[
\mathbf{v} = \mathbf{u} + \mathbf{a}t = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \cdot 10
\]

\[
\mathbf{v} = 3\hat{i} + 4\hat{j} + (4\hat{i} + 3\hat{j}) = (3 + 4)\hat{i} + (4 + 3)\hat{j} = 7\hat{i} + 7\hat{j}
\]

Calculate speed:

\[
\text{Speed} = |\mathbf{v}| = \sqrt{(7)^2 + (7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \, \text{m/s}
\]

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Question 5

A point initially at rest moves along x-axis. Its acceleration varies with time as a = (6t + 5) ms–². If it starts from origin, the distance covered in 2 s is

Answer

Given the acceleration:

\[
a = 6t + 5 \, \text{m/s}^2
\]

1. Integrate acceleration to find velocity $v$:

\[
v = \int a \, dt = \int (6t + 5) \, dt = 3t^2 + 5t + C
\]

Since it starts from rest, $C = 0$:

\[
v = 3t^2 + 5t
\]

2. Integrate velocity to find displacement $x$:

\[
x = \int v \, dt = \int (3t^2 + 5t) \, dt = t^3 + \frac{5}{2}t^2 + D
\]

Starting from the origin gives $D = 0$:

\[
x = t^3 + \frac{5}{2}t^2
\]

3. Calculate displacement at $t = 2$ s:

\[
x(2) = (2)^3 + \frac{5}{2}(2)^2 = 8 + \frac{5}{2} \cdot 4 = 8 + 10 = 18 \, \text{m}
\]

Thus, the distance covered in 2 seconds is: 18 m

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Question 6

If the velocity of a particle is given by \[v=(180-16x)^{1/2} ms^{-1}\], then its acceleration will be

Answer

Given the velocity:

$v = \sqrt{180 - 16 x} m/s$

To find the acceleration, use the chain rule:

$a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = \frac{d v}{d x} \cdot v$

Differentiate

$v$

$v$ with respect to

$x$

$x$ :

$\frac{dv}{dx} = \frac{-8}{\sqrt{180 - 16x}}$

Now, substitute

$v = \sqrt{180 - 16x}$

$a = - 8 {m/s}^{2}$

Thus, the acceleration is

$a = -8 \, \text{m/s}^2$

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Question 7

A bus begins to move with an acceleration of 1 ms–². A man who is 48 m behind the bus starts running at 10 ms–¹ to catch the bus. The man will be able to catch the bus after

Answer

Using relative speed, let’s re-calculate with the correct setup:

1. Acceleration of the bus: $ a = 1 \, \text{m/s}^2 $

2. Initial distance behind the bus: $ d = 48 \, \text{m} $

3. Speed of the man: $ v_m = 10 \, \text{m/s} $

4. The equation of motion for the bus after $ t $ seconds:
\[
d_b = \frac{1}{2} a t^2 = \frac{1}{2} \cdot 1 \cdot t^2 = \frac{1}{2} t^2
\]

5. Distance traveled by the man:
\[
d_m = v_m \cdot t = 10t
\]

6. Setting the distances equal considering the man starts 48 m behind:
\[
10t = \frac{1}{2} t^2 + 48
\]
\[
\frac{1}{2} t^2 - 10t + 48 = 0
\]

7. Multiplying by 2:
\[
t^2 - 20t + 96 = 0
\]

8. Using the quadratic formula:
\[
t = \frac{20 \pm \sqrt{20^2 - 4 \cdot 96}}{2}
\]
\[
t = \frac{20 \pm \sqrt{400 - 384}}{2}
\]
\[
t = \frac{20 \pm 4}{2}
\]
\[
t = 12 \, \text{s} \quad \text{or} \quad t = 8 \, \text{s}
\]

the man will catch the bus is 8 seconds.

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Question 8

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is :

Answer

Speed at mid point is given by

\[ V_{mid}=\sqrt{\frac{V_{1}^{2}+V_{2}^{2}}{2}} \]

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Question 9

Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1 : 8 : 1 and the maximum speed attained be 60 km h–¹, then what is the average speed over the whole journey?

Answer

Time Ratios: The time ratios are given as $1:8:1$

$1 : 8 : 1$ . So, if the total time is $t$

$t$ , the train spends $\frac{t}{10}$

$10 t$ accelerating, $\frac{8t}{10}$

$10 8 t$ at constant velocity, and $\frac{t}{10}$

$10 t$ decelerating.
Velocity-Time Graph:
- The graph forms a trapezium.
- The train accelerates linearly from 0 to 60 km/h, holds constant at 60 km/h, and then decelerates back to 0.
Key points:
- The area under this graph represents the total distance traveled.
- The height (maximum velocity) = 60 km/h.
- The time intervals are in the ratio $1:8:1$
  
  $1 : 8 : 1$ .
Average Speed:
The area of the trapezium is given by:

$\text{Area} = \frac{1}{2} \times (t_{\text{total}}) \times (\text{initial velocity} + \text{final velocity})$ $Area = 21 \times (t_{total}) \times (initial velocity + final velocity)$ For the constant velocity portion:

$\text{Average speed} = \frac{60 \times (1 + 8 + 1)}{10} = 54 \, \text{km/h}$ $Average speed = 1060 \times ( 1 + 8 + 1 ) = 54 km/h$

Thus, the average speed is 54 km/h.

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Question 10

A particle is thrown upwards from ground. It experience a constant air resistance force which can produce a retardation of 2 m/s². The ratio of time of ascent to the time of descent is : [g = 10 m/s²]

Answer

Effective accelerations:
- Ascent: $ a_{\text{eff}} = g + a = 10 + 2 = 12 \, \text{m/s}^2 $
- Descent: $ a'_{\text{eff}} = g - a = 10 - 2 = 8 \, \text{m/s}^2 $

Ratio of times:
\[
\frac{t_a}{t_d} = \sqrt{\frac{a'_{\text{eff}}}{a_{\text{eff}}}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}}
\]

The ratio of time of ascent to time of descent is √(2/3).

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