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(D) The car accelerates from rest to a maximum speed of $20\,m/s$ at $a = 1\,m/s^2$. The time taken for acceleration is $t_1 = v/a = 20/1 = 20\,s$. Th

Vedclass · Accepted Answer

$65$

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(D) The car accelerates from rest to a maximum speed of $20\,m/s$ at $a = 1\,m/s^2$. The time taken for acceleration is $t_1 = v/a = 20/1 = 20\,s$. The distance covered during this phase is $d_1 = (1/2) 	imes a 	imes t_1^2 = (1/2) 	imes 1 	imes 20^2 = 200\,m$.The car decelerates from $20\,m/s$ to rest at $a' = 2\,m/s^2$. The time taken for deceleration is $t_3 = v/a' = 20/2 = 10\,s$. The distance covered during this phase is $d_3 = (1/2) 	imes a' 	imes t_3^2 = (1/2) 	imes 2 	imes 10^2 = 100\,m$.The total distance is $1\,km = 1000\,m$. The distance covered at constant speed is $d_2 = 1000 - (200 + 100) = 700\,m$. The time taken at constant speed is $t_2 = d_2 / v = 700 / 20 = 35\,s$.The total time taken is $T = t_1 + t_2 + t_3 = 20 + 35 + 10 = 65\,s$.

$A$ small electric car has a maximum constant acceleration of $1\,m/s^2$,a maximum constant deceleration of $2\,m/s^2$,and a maximum speed of $20\,m/s$. The amount of time it would take to drive this car $1\,km$ starting from rest and finishing at rest is $.........\,s$.

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