NEET Kinematics MCQs & PYQs

Q: A man wishes to swim across a river 400 m wide flowing with a speed of 3m/s, such that he reaches the point just in front on the other bank in time not greater than 100 sec. The angle made by the direction he swims and river flow direction is :

Distance travelled in y-direction = 400 m Time to cross = 100 sec So, Speed in y-direction = 4 m/s Speed in x-direction = 3m/s So, angle = 90 +37 =127º

Q: Rain drops are falling vertically and a man is running on the ground horizontally with speed equal to the rain with respect to the ground. Rain drops appears to him hitting with velocity 10 km/hr. Find velocity of rain.

Velocity of Rain = - V j Velocity of Person = V i Velocity of Rain with respect to the Person = - V i - V j so, V√2 = 10 ⇒ V = 5√2 km/hr

Q: A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn/Sn+1 is :

Sn = u + a/2(2n-1) = a/2(2n-1) Sn+1 = u + a/2(2(n+1)-1) = a/2(2n+1) Sn / Sn+1 = (2n-1)/(2n+1)

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: A body moves 6 m north, 8 m east and 10 m vertically upwards, what is its resultant displacement from initial position?

1. Movement 1: 6 m North \[ \text{Displacement}_1 = 6 \hat{j} \] 2. Movement 2: 8 m East \[ \text{Displacement}_2 = 8 \hat{i} \] 3. Movement 3: 10 m Vertically Upwards \[ \text{Displacement}_3 = 10 \hat{k} \] Total Displacement: \[ \text{Total Displacement} = 6 \hat{j} + 8 \hat{i} + 10 \hat{k} \] Magnitude of Displacement: \[ |\text{Total Displacement}| = \sqrt{(8^2) + (6^2) + (10^2)} = \sqrt{64 + 36 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m} \]

Q: A person moves 30 m north and then 20 m towards east and finally 30√2 min south -west direction. The displacement of the person from the origin will be

1. Movement 1: 30 m North $ \text{Displacement}_1 = 30 \hat{j} $ 2. Movement 2: 20 m East $ \text{Displacement}_2 = 20 \hat{i} $ 3. Movement 3: $ 30\sqrt{2}$ m Southwest $ \text{Displacement}_3 = -30 \hat{i} - 30 \hat{j} $ Total Displacement: $ \text{Total Displacement} = (20 \hat{i} + 30 \hat{j}) + (-30 \hat{i} - 30 \hat{j}) = -10 \hat{i} + 0 \hat{j} $ Displacement = 10 m West

Question 1

A man walks in rain at 72 cm/s due east and observes the rain falling vertically. When he stops, rain appears to strikes his back at 37º from the vertical. Find velocity of raindrops relative to the ground.

Answer

💬 Discuss this Question

Question 2

A man wishes to swim across a river 400 m wide flowing with a speed of 3m/s, such that he reaches the point just in front on the other bank in time not greater than 100 sec. The angle made by the direction he swims and river flow direction is :

Answer

Distance travelled in y-direction = 400 m

Time to cross = 100 sec

So, Speed in y-direction = 4 m/s

Speed in x-direction = 3m/s

So, angle = 90 +37 =127º

💬 Discuss this Question

Question 3

Rain drops are falling vertically and a man is running on the ground horizontally with speed equal to the rain with respect to the ground. Rain drops appears to him hitting with velocity 10 km/hr. Find velocity of rain.

Answer

Velocity of Rain = - V j

Velocity of Person = V i

Velocity of Rain with respect to the Person = - V i - V j

so, V√2 = 10 ⇒ V = 5√2 km/hr

💬 Discuss this Question

Question 4

The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B ?

Answer

From Equations of motion

The height of a jump is inversely proportional to the acceleration due to gravity. Let $ h_A $ be the height of the jump on planet A, and $ h_B $ be the height of the jump on planet B. Also, let $ g_A $ and $ g_B $ represent the acceleration due to gravity on planets A and B, respectively.

Given:
- $ g_A = 9g_B $
- $ h_A = 2 \, \text{m} $

The ratio of heights is:

\[
\frac{h_B}{h_A} = \frac{g_A}{g_B} = 9
\]

Thus, the height on planet B is:

\[
h_B = 9 \times h_A = 9 \times 2 = 18 \, \text{m}
\]

So, the height of the jump on planet B is $ 18 \, \text{m} $.

💬 Discuss this Question

Question 5

A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = n – 1 to t = n. Then Sn/Sn+1 is :

Answer

S_n = u + a/2(2n-1) = a/2(2n-1)

S_n+1 = u + a/2(2(n+1)-1) = a/2(2n+1)

S_n / S_n+1 = (2n-1)/(2n+1)

💬 Discuss this Question

Question 6

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).

💬 Discuss this Question

Question 7

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t – 5t²) meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is :

Answer

Given:

\[
y = 8t - 5t^2 \quad \text{and} \quad x = 6t
\]

To find the initial velocity, we calculate the horizontal and vertical components of velocity:

1. Horizontal velocity $ v_x = \frac{dx}{dt} = 6 \, \text{m/s} $
2. Vertical velocity $ v_y = \frac{dy}{dt} = 8 - 10t $

At $ t = 0 $ (initial velocity):

\[
v_y(0) = 8 \, \text{m/s}
\]

Thus, the magnitude of the initial velocity $ u $ is:

\[
u = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s}
\]

The velocity of projection is $ 10 \, \text{m/s} $.

💬 Discuss this Question

Question 8

A body moves 6 m north, 8 m east and 10 m vertically upwards, what is its resultant displacement from initial position?

Answer

1. Movement 1: 6 m North
\[
\text{Displacement}_1 = 6 \hat{j}
\]

2. Movement 2: 8 m East
\[
\text{Displacement}_2 = 8 \hat{i}
\]

3. Movement 3: 10 m Vertically Upwards
\[
\text{Displacement}_3 = 10 \hat{k}
\]

Total Displacement:
\[
\text{Total Displacement} = 6 \hat{j} + 8 \hat{i} + 10 \hat{k}
\]

Magnitude of Displacement:

\[
|\text{Total Displacement}| = \sqrt{(8^2) + (6^2) + (10^2)} = \sqrt{64 + 36 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m}
\]

💬 Discuss this Question

Question 9

A person moves 30 m north and then 20 m towards east and finally 30√2 min south -west direction. The displacement of the person from the origin will be

Answer

1. Movement 1: 30 m North $ \text{Displacement}_1 = 30 \hat{j} $

2. Movement 2: 20 m East $ \text{Displacement}_2 = 20 \hat{i} $

3. Movement 3: $ 30\sqrt{2}$ m Southwest $ \text{Displacement}_3 = -30 \hat{i} - 30 \hat{j} $

Total Displacement: $
\text{Total Displacement} = (20 \hat{i} + 30 \hat{j}) + (-30 \hat{i} - 30 \hat{j}) = -10 \hat{i} + 0 \hat{j} $

Displacement = 10 m West

💬 Discuss this Question

Question 10

A car moves a distance of 200 m. It covers first half of the distance at speed 60 kmh–¹ and the second half at speed v. If the average speed is 40 kmh–¹, the value of v is

Answer

\[ V_{av}= \frac{2V_{1}.V_{2}}{V_{1}+V_{2}}\]

\[ 40= \frac{2\times 60\times V_{2}}{ 60+V_{2}} \]

\[ V_{2}= 30 m/s \]

💬 Discuss this Question

NEET Kinematics MCQs & PYQs

Practice NEET Kinematics Questions