Question

a car moving with a uniform velocity of 44 metre per second overtakes and the car be which is just moving from rest with an acceleration of 2 metre per second after how much time do they meet again

Answer 1

The second car should gain the constant velocity of first car which is 44m/s
So, final velcity (v)= 44m/s
initial velocity (u)=0
taken time=t s
acceleration (a) =2m/s^2
Using formula,
v=u+at
=>44=0+2t
or, t=44/2=22s

Ans. 22s


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Answer 2

Here's your answer..

Solution:-
Initial velocity(u) =0
Final Velocity(v) =44m/s
Acceleration=2m/s^2
Time =?

We will find time by using equation:-
$= > v = u + at \\ = > 44 = 0 + 2t \\ = > 44 = 2t$
$= > t = \frac{44}{2} \\ = > t = 22$
Hence,
Time taken is 22 seconds.

Hope it helps you....