A car starts from rest and moves along the x- | vedclass

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Question

(A) The distance travelled in the first $8 \, s$ is calculated using the equation $s = ut + \frac{1}{2}at^2$. Since the car starts from rest,$u = 0$.$

Vedclass · Accepted Answer

$320$

Vedclass · Answer

(A) The distance travelled in the first $8 \, s$ is calculated using the equation $s = ut + \frac{1}{2}at^2$. Since the car starts from rest,$u = 0$.$x_1 = 0 + \frac{1}{2} 	imes 5 	imes (8)^2 = 160 \, m$.At $t = 8 \, s$,the velocity $v$ is calculated as $v = u + at = 0 + (5 	imes 8) = 40 \, m/s$.For the remaining time,which is $12 \, s - 8 \, s = 4 \, s$,the car moves with a constant velocity of $40 \, m/s$.The distance covered in the last $4 \, s$ is $x_2 = v 	imes t = 40 	imes 4 = 160 \, m$.Therefore,the total distance covered in $12 \, s$ is $x = x_1 + x_2 = 160 \, m + 160 \, m = 320 \, m$.

$A$ car starts from rest and moves along the $x-$ axis with constant acceleration $5\, ms^{-2}$ for $8\, s$. If it then continues with constant velocity,what distance will the car cover in $12\, s$ since it started from the rest (in $m$)?

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