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A truck travelling along a straight road with a constant speed of 72 km/hr passes a 0 moving much slower. At the instant the truck passes the car, the car at time t car starts accelerating with constant acceleration 1m/s and overtakes the truck 0.6km further down the road The car moves with uniform velocity from this instant onwards. Find the distance between them at time t 50 sec

```
4 years ago

```							Since in the question it is not mentioned about the initial velocity of the car, let it be at rest. The velocity of truck,  .   VT = 20 m/s. Let the meet after t seconds.Initial velocity of car, uC = 0, acceleration, a = 1m/s2 Distance travelled in t sec, S1 = at2/2Distance travelled by truck, S2 = vTtThen they are equal.So, 20.t = 1.t2/2     => t=40 s Velocity attained by the csr in 40 s, vc = at = 1.40 = 40 m/s.From hence, relative velocity if car w.r.t the truck, VCT = vc - vT = 40 - 20 = 20 m/s.So from the starting, distance between them is VCT.t = 20*30 = 600 m
```
4 years ago
```							Or otherwise, let the initial velocity of the car be u. The velocity of truck,  .   VT = 20 m/s. Let the car takes over the truck after t secondsIn this time,Distance travelled by truck, S2 = vTt     => 600 = 20.t      => t = 30 sSo, the car takes over the truck after 30 seconds  And, distance travelled by the car, S = u.t + at2/2     => 600 = 30.t + 302/2.    => u = 5 m/s. Velocity attained by the csr in 30 s, vc = u + at = 5 + 1.30 = 35 m/s.From hence, relative velocity if car w.r.t the truck, VCT = vc - vT = 35 - 20 = 15 m/s.So, at t = 50 s, the car is ahead of the truck for (50-30) = 20 sHence, the distance between them is X = VCT.20 = 15*20 = 300 mTHANKS
```
4 years ago
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