A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be

Question

A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be (A) ((α2+β2/α β)) t (B) ((α2-β2/α β)) t (C) ((α+β/α β)) t (D) ((α β/α+β)) t

Tardigrade Admin · Accepted Answer

Initial velocity (u)=0; Acceleration in the first phase =α; Deceleration in the second phase =β and total time =t. When car is accelerating then final velocity (v)=u+α t=0+α t1 or t1=(v/α) and when car is decelerating, then final velocity 0=v-β t or t2=(v/β). Therefore total time (t)=t1+t2=(v/α)+(v/β) t=v((1/α)+(1/β))=v((β+α/α β)) or v=(α β t/α+β).

Q. A car accelerates from rest at a constant rate $α$ for some time after which it decelerates at a constant rate $β$ and comes to rest. If total time elapsed is $t$ , then maximum velocity acquired by car will be

$(\frac{α ^{2} + β ^{2}}{α β}) t$

$(\frac{α ^{2} - β ^{2}}{α β}) t$

$(\frac{α + β}{α β}) t$

$(\frac{α β}{α + β}) t$

Q. A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be

(αβα2+β2​)t

(αβα2−β2​)t

(αβα+β​)t

(α+βαβ​)t

Q. A car accelerates from rest at a constant rate $α$ for some time after which it decelerates at a constant rate $β$ and comes to rest. If total time elapsed is $t$ , then maximum velocity acquired by car will be

$(\frac{α ^{2} + β ^{2}}{α β}) t$

$(\frac{α ^{2} - β ^{2}}{α β}) t$

$(\frac{α + β}{α β}) t$

$(\frac{α β}{α + β}) t$