A lift starts going up from rest. It acceleration is shown in figure and which is plotted against time. The time in which the lift comes to instantaneous rest is :

4 s

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8 s

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10 s

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12 s

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Solution

The correct option is D $12 s$
velocity at t=4secs $v = a t = 4$ m/s
from t=4secs to t=8secs acc=0 therefore velocity is 4m/s
velocity at t=12secs $v = 4 - 4 \times 1 = 0$
hence amswer is $12 s$

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Q. A lift ascends from rest with uniform acceleration of

4 m / s^{2},

then it moves with uniform velocity and finally comes to rest with uniform retardation of

4 m / s^{2} .

If the total distance covered during ascending is

28 m

and total time taken for ascending is

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respectively

.

Find the time for which the lift moves with uniform velocity

.

Also find its uniform velocity

.

A block of mass 12 kg is hanging with a light string in a lift as shown in figure (a). The lift is moving in the upward direction and its speed time graph is plotted as shown in figure (b) : (take

g = 10 m / s^{2}

)

Which of the following is possible?

A lift with open top is moving up with acceleration $f$ . A body is thrown up with speed $u$ relative to lift and it comes back to lift in time. Then speed $u$ is (g=acceleration due to gravity)

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A lift starts going up from rest. It acceleration is shown in figure and which is plotted against time. The time in which the lift comes to instantaneous rest is :

The correct option is D 12 svelocity at t=4secs v=at=4m/sfrom t=4secs to t=8secs acc=0 therefore velocity is 4m/svelocity at t=12secs v=4−4×1=0hence amswer is 12s

The correct option is D $12 s$
velocity at t=4secs $v = a t = 4$ m/s
from t=4secs to t=8secs acc=0 therefore velocity is 4m/s
velocity at t=12secs $v = 4 - 4 \times 1 = 0$
hence amswer is $12 s$