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15.A car travelling at 15m/s comes to rest due to application of brakes that producea deceleration of 5 m/s2. The stopping distance isa) 1.5mb) 45mc) 13.5m / d) 22.5m |
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Answer» Using,v = u + at v = 0u = 15 m/sa = - 5 m/s^2 0 = 15 - 5*tt = 15/5 = 3 d Now, usings = ut + 1/2 a t^2 = 15*3 + 1/2 * (-5) * 3*3 = 45 - 45/2 = (90 - 45)/2 = 45/2 = 22.5 m (d) is correct option |
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