Q10 / 103 · NTA Abhyas (NEET) 2020 · Physics · Motion In One Dimension

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 is

  • A
    ( α 2 − β 2 αβ )t
  • B
    ( α 2 + β 2 αβ )t
  • C
    (α+β)t αβ
  • Correct answer
    D
    αβt α+β

Solution

Let us draw the   v−t  graph of the given situation. Initially it accelerates uniformly from rest. Hence, it is a straight line.

Figure

    v max =0+α t 1 ⇒   t 1 = v max α       0= v max −β t 2 ⇒ t 2 = v max β

In Figure

t=t1+t2=vmaxα+vmaxβ

t=vmax1α+1β=vmaxα+βαβ

vmax=tαβα+β

Hence, the correct answer is option: 'D'.

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