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. If total time elapsed is t, then maximum velocity acquired by car will be ( 1994 )

{ (\\alpha^{2}-\\beta^{2} ) t}/{\\alpha \\beta}

{ (\\alpha^{2}+\\beta^{2} ) t}/{\\alpha \\beta}

\frac{(\\alpha+\\beta) t}{\\alpha \\beta}

\frac{\\alpha \\beta t}{\\alpha+\\beta}

Class 11 NEET Physics Motion in a Straight Line Questions

Question : 49 of 60

Marks: +1, -0

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

( 1994 )

(α2−β2)t
αβ
(α2+β2)t
αβ
(α+β)t
αβ
αβt
α+β

Validate

Go to Question:

Prev Question

Next Question