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A truck is moving at a speed of 50 ms-1 Find stopping distance if it retards at a rate of 20 m/min. |
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Answer» tion:ANSWERAvg SPEED × total times = total DISTANCE 20×25=500Since ACCELERATION in the 1st interval is same as deceleration in the 3rd interval and deceleration is from the same velocity ACHIEVED after acceleration, t 1 =t 3 s 1 = 21 ×5t 12 s 2 =(5t 1 )t 2 s 3 =(5t 1 )×t 3 − 21 ×5×t 12 =(5t 1 )×t 1 − 21 ×5×t 12 s 1 +s 2 +s 3 =5t 1 t 2 +5t 12 ⇒5t 12 +5t 1 t 2 =500−−−−12t 1 +t 2 =25−−−−2Solving (1 ) and (2) ⇒t 1 =5,t 2 =15sec |
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