Question

(23) A man is at a distance of 6 m from a bus.
The bus begins to move with a constant
acceleration of 3 m/s2. In order to catch
the bus, the minimum speed with which
the man should run towards the bus is ...
[AIIMS-2016]
(A) 2 m/s
(B) 4 m/s
(C) 6 m/s
(D) 8 m/s​

Answer 1

Required Solution:

Option C

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How?

Given:

• Distance between man and bus (s) = 6m
• Acceleration of the bus (a) = $\sf {3m/{s}^{2}}$

To calculate:

• To catch the bus, the minimum speed with which the man should run towards the bus?

Solution:

Here, we can solve it by using 3rd equation of motion. Final velocity of the bus will be the minimum speed with which the man should run towards the bus.

So, we know that,

• $\sf { 2as = {v}^{2}-{u}^{2}}$

Here, initial velocity(u) will be 0 as he starts from rest to catch the bus.

$\sf {\longrightarrow 2 \times 3 \times 6 = {v}^{2}-{0}^{2}}$

$\sf {\longrightarrow 36 = {v}^{2} }$

$\sf {\longrightarrow \sqrt{36} = v }$

$\sf {\longrightarrow 6 = v }$

Therefore, final velocity of the man will be 6m/s or we can say that the minimum speed with which the man should run towards the bus is 6m/s.

Thus, option C is correct.

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Quick check!

Obviously we got the answer that is 6m/s but let's verify it too.

◆ In the given scenario, that man has to cover 6 m to catch the bus. Now, suppose that his final velocity is 6m/s. So, now if he is travelling with the speed of 6m/s, then he would cover how much distance?

If we get the value of distance as 6m. So our answer is correct.

→ By the 3rd third equation of motion:

• $\sf { 2as = {v}^{2}-{u}^{2}}$

$\sf {\longrightarrow 2 \times 3 \times s = {6}^{2}-{0}^{2}}$

$\sf {\longrightarrow 6 \times s = {6}^{2}}$

$\sf {\longrightarrow s = \dfrac{36}{6} }$

$\sf {\longrightarrow s = 6m}$

so, if he travels with the speed of 6m/s , he will cover the distance of 6m as the distance between the bus and him is 6m.

Hence verified!

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