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A car is travelling on straight level road with a uniform speed of 60kmph it is followed by another car B which is moving with speed of 70kmph when the distance between them is 2.5km the car Back is given a deceleration of 20km/h2 after what distance of time B will catch up with A
Suppose the car B catches up with the car A in t hours. The distance travelled by the car Amoving with a velocity of 60 km h-1 in time hours is given by s1 = 60 t km.The distance traveled by the car B moving with an initial velocityu = 70 km h-1 and decelerated at the rate of 20 km h-2 in time t hours is given bys2 = ut + ½ at2 = 70t + ½ (-20)t2 = 70t - 10t2 But (s2-s1) = 2.5 km.·. 70 t -10 t2 - 60 t = 2.5 => 10t2 - 10t + 2.5 = 0.·. t2 - t + 0.25 = 0 => (t - 0.5)2 = 0=> t = ½ hSubstituting t = ½ h, we get s2 = 32.5 km.
s2 = ut + ½*a*t2 = 70t + ½*(-20)*t2 = 70t -10t2s1 = 60ts2 – s1 = 2.570t – 10t2 -60t = 2.510t2 -10t + 2.5 = 0so t = ½ hputting this in s2s2 = 32.5kmHope it clears.
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