First slide

Rectilinear Motion

Question

A lift performs the first part of ascent with uniform acceleration a and the remainder with uniform retardation 2a. The lift starts from rest and finally comes to rest. If t is the time of ascent. Find the height ascended by lift.

Moderate

A

$\frac{a t^{2}}{4}$
B

$\frac{a t^{2}}{3}$
C

$\frac{a t^{2}}{2}$
D

$\frac{a t^{2}}{8}$

Solution

Let OAB be the velocity-time graph of the lift. The ordinate at A (i.e., AM) represents maximum velocity. Total distance travelled will be
$\begin{array}{l} Area of the Δ O A B = \frac{1}{2} \times O B \times A M \\ A M = v, O M = t_{1}, t_{1} + t_{2} = O B = t, M B = t_{2} \\ Δ O A B = \frac{1}{2} \times t v = h \\ or v t = 2 h . . . . . . . (i) \\ Now \frac{v}{t_{1}} = a or t_{1} = \frac{v}{a} . . . . . . . . (i i) \\ and \frac{v}{t_{2}} = 2 a or t_{2} = \frac{v}{2 a} . . . . . . . . . (i i i) \end{array}$
Adding (ii) and (iii)
$\begin{array}{l} t = t_{1} + t_{2} = \frac{v}{a} + \frac{v}{2 a} = \frac{3 v}{2 a} = \frac{3}{2 a} \times \frac{2 h}{t} \\ or a t^{2} = 3 h \Rightarrow h = \frac{a t^{2}}{3} \end{array}$

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