A car accelerates form rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by

Question

A car accelerates form rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by (A) (α β/α+β) t (B) (α+β/α β) t (C) (α2+β2/β α) t (D) (α2-β2/β α) t

Tardigrade Admin · Accepted Answer

Let t1 be the time during which the car accelerates at a rate α. The velocity at the end of time t1 will be v at t1=u+α t1=0+α t1=α t1 (∵ u=0) The time during which the car decelerates is t2=t-t1. For this time t2 prime the initial velocity is α t1 and the final velocity is zero and the acceleration is -β Therefore 0=α t1-β(t-t1) which gives t1=(β t/α+β) . Therefore, maximum velocity =v at t1=α t1=(α β t/α+β)

Q. A car accelerates form rest at a constant rate $α$ for some time after which it decelerates at a constant rate $β$ to come to rest. If the total time elapsed is $t$ , the maximum velocity acquired by the car is given by

$\frac{α β}{α + β} t$

$\frac{α + β}{α β} t$

$\frac{α ^{2} + β ^{2}}{β α} t$

$\frac{α ^{2} - β ^{2}}{β α} t$

Q. A car accelerates form rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by

α+βαβ​t

αβα+β​t

βαα2+β2​t

βαα2−β2​t

Q. A car accelerates form rest at a constant rate $α$ for some time after which it decelerates at a constant rate $β$ to come to rest. If the total time elapsed is $t$ , the maximum velocity acquired by the car is given by

$\frac{α β}{α + β} t$

$\frac{α + β}{α β} t$

$\frac{α ^{2} + β ^{2}}{β α} t$

$\frac{α ^{2} - β ^{2}}{β α} t$