A car accelerates from rest at a constant rat | vedclass

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(B) Let the acceleration be $\alpha = 2 \ m/s^2$ and retardation be $\beta = 4 \ m/s^2$. The total time is $t = 3 \ s$.Let $t_1$ be the time of accele

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$6$

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(B) Let the acceleration be $\alpha = 2 \ m/s^2$ and retardation be $\beta = 4 \ m/s^2$. The total time is $t = 3 \ s$.Let $t_1$ be the time of acceleration and $t_2$ be the time of retardation. Then $t_1 + t_2 = 3$.Since the car starts from rest and comes to rest,the maximum velocity $v$ reached is $v = \alpha t_1 = \beta t_2$.Substituting the values,$2t_1 = 4t_2$,which implies $t_1 = 2t_2$.Substituting into the time equation: $2t_2 + t_2 = 3$,so $3t_2 = 3$,which gives $t_2 = 1 \ s$ and $t_1 = 2 \ s$.The maximum velocity is $v = 2 	imes 2 = 4 \ m/s$.The total displacement $d$ is the area under the velocity-time graph,which is a triangle: $d = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 4 = 6 \ m$.

$A$ car accelerates from rest at a constant rate of $2 \ m/s^2$ for some time. Then it retards at a constant rate of $4 \ m/s^2$ and comes to rest. If it remains in motion for $3 \ s$,then the total displacement covered by it is $.... \ m$.

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