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Motion in a Straight Line

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Question : 7 of 27
Marks: +1, -0
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h−1^{-1} in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s−2^{-2}. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Solution:  
For train A:A :  uA=72 km h−1=20 ms−1;\; u_{A}=72\ \text{km}\ \text{h}^{-1}=20\ \text{ms}^{-1};
t=50 s;a=0t=50\ \text{s}; a=0
∴  SA=uAt\therefore\; S_{A}=u_{A} t
or SA=20×50S_{A}=20 \times 50
=1,000 m=1{,}000\ \text{m}
For train B:B :  uB=72 km h−1\; u_{B}=72\ \text{km}\ \text{h}^{-1}
=20 m s−1=20\ \text{m}\ \text{s}^{-1}
t=50 s;a=1 m s−2t=50\ \text{s}; a=1\ \text{m}\ \text{s}^{-2}
∴  SB=uBt+12at2\therefore\; S_{B}=u_{B} t+\frac{1}{2} a t^{2}
or SB=20×50+12×1×(50)2S_{B}=20 \times 50+\frac{1}{2} \times 1 \times (50)^{2}
=2250 m=2250\ \text{m}
Let the original distance between the two trains be S.
S=SB−SAS=S_{B}-S_{A}
or S=2,250−1,000S=2{,}250-1{,}000
=1250 m=1250\ \text{m}
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