Question

Two cars A and B are at rest at same point initially. If a starts with uniform velocity of 40 metre per second and B starts in same direction with constant acceleration of 4 metre per second square ,then B will catch A after how much time ?

Answer 1

Here's your answer!!

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Let the distance cover by A be "d1"
And Distance Covered by B be "d2"

In case of Car "A"

Initial velocity of A =40m/s
Time = t
Acceleration= 0m/s^2

Now,
In case of Car "B" :-
Initial velocity= 0
Time =t
Acceleration= 4m/s^2

We know that,
$= > s = ut \frac{1}{2} {at}^{2}$
B will catch A when D1=D2

Therefore,

D1=D2
$ut + \frac{1}{2} a {t}^{2} = ut + \frac{1}{2} {at}^{2}$
$40 \times t + \frac{1}{2} \times 0 \times {t}^{2} = 0 \times t + \frac{1}{2} \times 4 \times {t}^{2}$
$40t + \frac{1}{2} \times 0 \times {t}^{2} = 0t + \frac{1}{2} \times 4 \times {t}^{2} \\ = 40t = 2 {t}^{2} \\ = {t}^{2} = 20t \\ = t = 20$

Hence,
Time taken by B to catch A is 20 seconds.

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Hope it helps you !! :)