First slide

Instantaneous velocity

Question

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 = 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 vt = 2h ..(i)
$N o w \frac{v}{t_{1}} = a o r t_{1} = \frac{v}{a} ..... (i i)$
$a n d \frac{v}{t_{2}} = 2 a o r t_{2} = \frac{v}{2 a} ..... (i i i)$
Adding (ii) and (iii)
$t = t_{1} + t_{2} = \frac{v}{a} + \frac{v}{2 a} = \frac{3 v}{2 a} = \frac{3}{2 a} \times \frac{2 h}{t}$
$o r a t^{2} = 3 h \Rightarrow h = \frac{a t^{2}}{3}$

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