At time t = 0, Car A, which is travelling at a constant speed of 20 ms–1 on a straight horizontal road, overtakes Car B travelling at a speed of 15 ms–1. Car B immediately accelerates uniformly and, T seconds later, it overtakes Car A, which has kept its speed at 20 ms–1. The distance travelled by each car in time T is 600 m.

Find the speed of car B at the moment it overtakes car A.

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Since speed of the car remains unchanged 20ms^(-1)

Car A travelld 600 m in T sec

Apply equation linear motion

`s=ut+(12)at^2`

since speed remains same so , a=0 ,Thus

600=20 x T

T=30 sec.

Acceleration for car B

`600=15xx30+(1/2)a (30)^2`

`600=450+(900/2)a`

`600-450=450a`

`150/450=a`

`a=(1/3)ms^(-2)`

Thus when car B was overtaking car A ,its speed will be determined by third equation of motion

`v^2=u^2+2as`

`v^2=(15)^2+2(1/3)xx600`

`v^2=225+400`

`v^2=625`

`v=sqrt(625)`

`v=25ms^(-1)`

Thus when car B overtook car A ,its speed was 25m/s

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