A truck starting from rest moves with an acce | vedclass

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(C) The truck accelerates at $a_t = 5 \, m/s^2$ for $t = 1 \, s$. The maximum velocity attained by the truck is $v_{max} = a_t 	imes t = 5 	imes 1 =

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(C) The truck accelerates at $a_t = 5 \, m/s^2$ for $t = 1 \, s$. The maximum velocity attained by the truck is $v_{max} = a_t 	imes t = 5 	imes 1 = 5 \, m/s$.For the block of mass $m$ on the truck,the maximum acceleration it can have due to friction is $a_{max} = \mu g = 0.2 	imes 10 = 2 \, m/s^2$.Since the truck's acceleration $(5 \, m/s^2)$ is greater than the block's maximum possible acceleration $(2 \, m/s^2)$,the block will slip on the truck.The block will accelerate at $a_{max} = 2 \, m/s^2$ until its velocity matches the truck's velocity $(5 \, m/s)$.Let $t'$ be the time taken for the block to reach $5 \, m/s$. Then $v = a_{max} 	imes t' \implies 5 = 2 	imes t' \implies t' = 2.5 \, s$.Thus,the velocity of the block increases linearly with acceleration $2 \, m/s^2$ for $2.5 \, s$ until it reaches $5 \, m/s$,after which it moves with constant velocity. This corresponds to the graph in option $C$.

$A$ truck starting from rest moves with an acceleration of $5 \, m/s^2$ for $1 \, s$ and then moves with constant velocity. The velocity $w.r.t$ ground $v/s$ time graph for the block in the truck is (Assume that the block does not fall off the truck and the coefficient of friction $\mu = 0.2$):

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