Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DEasy2 minPYQ_2020

PhysicsEasynumerical

A lift ascends with constant acceleration $a = 1 m s^{- 2}$ , then with constant velocity and finally, it stops under constant retardation $a = 1 m s^{- 2}$ . If total distance ascended by the lift is $7 m$ , in a total time of the journey is $8 s$ . Find the time (in second) for which lift moves with constant velocity.

Answer:

6.00

Solution:

$a = \frac{v}{t_{1}}$

$t_{1} = \frac{v}{a}$ ...(i)

$t_{0} = 8 - 2 t_{1}$ ...(ii)

$s_{0} = 7 - 2 s_{1}$ ...(iii)

Solving equation (i), (ii) and (iii), we get

$t_{0} = 6 s$

Stream:NTA_ABHYASSubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...