Two cars A and B cross a point P with velocities 10/m and 15 m/s. After that, they move with different uniform accelerations and the car A overtakes B with a speed of 25 ms^-1. What is velocity of B at that instant?

Q. Two cars A and B start from the same starting point. A is starting from rest and moving with constant acceleration

2

m / s^{2}

and car B is starting with a constant velocity

40

m/s. After what time both cars will meet each other?

Q. Two cars, initially at a separation of

12 m

, start simultaneously. First car

A

, starting from rest moves with an acceleration

2 {m/s}^{2}

, whereas the car

B

, which is ahead, moves with a constant velocity

1 m/s

, away from car

A

along the same direction. Find the time when car

A

overtakes car

B

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Two cars A and B move such that car A moving with a uniform velocity of 15 ms−1 overtakes car B starting from rest with an acceleration of 3 ms−2. After how much time do they meet again?

The correct option is D 10sLet us assume that the car A and B meet after a time ′t′Distance travelled by A in that time is vt=15tDistance travelled by B iss=ut+12at2s=3t223t22=15tt=0,10st=0 is trivial . Therefore, at t=10s, both the cars meet.

Two cars A and B move such that car A moving with a uniform velocity of $15 m s^{- 1}$ overtakes car B starting from rest with an acceleration of $3 m s^{- 2}$ . After how much time do they meet again?