Starting from rest,a scooter acquires a veloc | vedclass

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Question

(A) Part $1$: Acceleration phaseInitial velocity $u = 0 \, m/s$,Final velocity $v = 36 \, km/h = 36 	imes (5/18) = 10 \, m/s$,Time $t_1 = 10 \, s$.Ac

Vedclass · Accepted Answer

Acceleration = $1 \, m/s^2$,Deceleration = $-0.5 \, m/s^2$,Distance = $150 \, m$

Vedclass · Answer

(A) Part $1$: Acceleration phaseInitial velocity $u = 0 \, m/s$,Final velocity $v = 36 \, km/h = 36 	imes (5/18) = 10 \, m/s$,Time $t_1 = 10 \, s$.Acceleration $a_1 = (v - u) / t_1 = (10 - 0) / 10 = 1 \, m/s^2$.Distance $s_1 = ut_1 + 0.5 a_1 t_1^2 = 0 + 0.5 	imes 1 	imes 10^2 = 50 \, m$.Part $2$: Deceleration phaseInitial velocity $u = 10 \, m/s$,Final velocity $v = 0 \, m/s$,Time $t_2 = 20 \, s$.Acceleration $a_2 = (v - u) / t_2 = (0 - 10) / 20 = -0.5 \, m/s^2$.Distance $s_2 = ut_2 + 0.5 a_2 t_2^2 = 10 	imes 20 + 0.5 	imes (-0.5) 	imes 20^2 = 200 - 100 = 100 \, m$.Total distance = $s_1 + s_2 = 50 + 100 = 150 \, m$.

Starting from rest,a scooter acquires a velocity of $36 \, km/h$ in $10 \, s$ and then brakes are applied,taking $20 \, s$ to stop. Calculate the acceleration and the total distance travelled.

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