A lift ascends with constant acceleration a = 1m s ^ - 2 , then...

A lift ascends with constant acceleration

a = 1m{s}^{- 2}

, then with constant velocity and finally, it stops under constant retardation

a = 1m{s}^{- 2}

. If total distance ascended by the lift is

7m

, in a total time of the journey is

8s

. Find the time (in second) for which lift moves with constant velocity. A lift ascends with constant acceleration a = 1m s ^ - 2 , then with constant velocity and finally, it stops under

A lift ascends with constant acceleration a = 1m s ^ - 2 , then with constant velocity and finally, it stops under

Correct answer

Step-by-step solution

a = \dfrac{v}{{t}_{1}}

{t}_{1} = \dfrac{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} = 6s

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