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 the total time elapsed is t, then the maximum velocity acquired by the car will be

Question

Accepted Answer

(4) (αβ t)/(α+β). Given: from rest, accelerate at α to a peak speed v, then decelerate at β to rest, total time t. Idea: write each phase's duration in terms of the same peak speed v, then add them to the given total. Speeding up: v=α t₁, so t₁=v/α. Slowing down: 0=v-β t₂, so t₂=v/β. t=t₁+t₂=v(1/α+1/β)=v(α+β)/(αβ). v=(αβ t)/(α+β).

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 the total time elapsed is t, then the maximum velocity acquired by the car will be

Step-by-step solution

Why the other options are wrong