The driver of a train A moving with a uniform speed of 30 ms^{−1} sees another train B moving with uniform speed 10 ms^{−1}

The driver of a train A moving with a uniform speed of 30 ms−1 sees another train B moving with uniform speed 10 ms−1 on the same track in the same direction. He immediately applies brakes and achieves a uniform retardation of 2ms−1 and finally stops. To avoid collision the minimum distance between the trains must be(a) 140 m(b) 75 m(c) 80 m(d) 120 m

Answer is : (b) 75 m The distance of A is given by \(s_A=ut+\frac{1}{2}at^2\) \(=30t+\frac{1}{2}(-2)t^2\) = 30t - t 2 and of B, s B = 10t From figure, s A - s B = s min ⇒ s min = 20t - t 2  …(i) Thus for A, v = u + at ⇒ 0 = 30 - 2t ⇒ t = 15 s Putting the value of t in Eq. (i), we get ∴ s min = 20 x 15 - 225 = 75 m

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