Question

Acceleration(a) versus time(t) curve for a lift is plotted below. find the height above its starting point where it comes to rest if it comes if it starts from rest .
1)20m
2)40m
3)60m
4)80m.

only correct answers accepted. ​

Answer 1

see the attachment for the answer

Answer 2

case 1 : time interval 0 to 2s

acceleration, a = 5m/s²

we know, acceleration is the rate of change of velocity with respect to time.

so, a = dv/dt = 5 ⇒v = 5t

so, velocity time graph is straight line.

at t = 2s , v = 5(2) = 10m/s

so, displacement = area under velocity-time graph

= 1/2 × 10m/s × 2s = 10m

case 2 : time interval 2s to 6s

acceleration = 0.

so, v = constant = 10m/s [ see figure]

so, displacement = area under velocity-time graph = area of rectangular part = 10m/s × (6s - 2s) = 40m

case 3 : time interval 6s to 8s

acceleration = -5m/s²

so, velocity, v = - 5t

so, displacement =1/2 × -5(8s-6s) × (8s - 6s) = -10m

now, net displacement = 10m + 40m - 10m = 40m

so, option (2) is correct choice.