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(A) The velocity-time graph is a trapezoid. Let the car accelerate for time $t_1$ and decelerate for time $t_1$. The time for which it moves with unif

Vedclass · Accepted Answer

$15$

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(A) The velocity-time graph is a trapezoid. Let the car accelerate for time $t_1$ and decelerate for time $t_1$. The time for which it moves with uniform velocity is $(25 - 2t_1) \ s$.Maximum velocity $v_{max} = a 	imes t_1 = 5t_1$.Average velocity $v_{avg} = 72 \ km \ hr^{-1} = 72 	imes \frac{5}{18} \ m \ s^{-1} = 20 \ m \ s^{-1}$.Total displacement $S = v_{avg} 	imes t_{total} = 20 	imes 25 = 500 \ m$.The area under the velocity-time graph is the displacement:$S = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$$500 = \frac{1}{2} 	imes [ (25 - 2t_1) + 25 ] 	imes 5t_1$$1000 = (50 - 2t_1) 	imes 5t_1$$1000 = 250t_1 - 10t_1^2$$10t_1^2 - 250t_1 + 1000 = 0$$t_1^2 - 25t_1 + 100 = 0$$(t_1 - 20)(t_1 - 5) = 0$Since $t_1$ must be less than $12.5 \ s$ (as $2t_1 The time for which the car moved with uniform velocity is $25 - 2t_1 = 25 - 2(5) = 15 \ s$.

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