Let after time t sect \space sect sec ,van catches up the car.
(a) So,distance traveled by car in ttt seconds:
ucart=uvant+12at2;uvan=0u_{car}t=u_{van}t+\frac{1}{2} at^2 ;u_{van}=0ucart=uvant+21at2;uvan=0
15t=12×a×t215 t=\frac{1}{2}\times a \times t^215t=21×a×t2
a=3m/sec2a=3 m/sec ^2a=3m/sec2
Solving this equation :
15t=12×3t215 t=\frac{1}{2}\times 3 t^215t=21×3t2
t=10 sect=10 \space sect=10 sec
(b) Distance traveled by car in 10 sec=10 \space sec =10 sec= ucart=15×10=150m\space u_{car}t=15 \times 10=150 m ucart=15×10=150m
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