Question

A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ and comes to rest. It total time elapsed is $$t$$, then maximum velocity acquired by car will be:

A
(α2β2)tαβ
B
(α2+β2)tαβ
C
(α+β)tαβ
D
(αβ)tα+β

Solution

The correct option is D $$\cfrac { \left( { \alpha }^{ }{ \beta }^{ } \right) t }{ \alpha +\beta }$$From rest, the car acceleraters for time t,So $$\alpha = \dfrac{V}{t_1}$$       ...(1) , let maximum velocity is V.The car begins decelerate after attaining maximum velocity$$\beta = \dfrac{V}{t-t_1}$$            ....(2)By adding 1 and 2$$t = v(\frac{1}{\alpha} + \dfrac{1}{\beta}$$$$v = \dfrac{(\alpha \beta)t}{\alpha + \beta}$$Hence (D) is correct answerPhysics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More