Question

An object which is initially at rest,accelerates at a rate of 4 m/s^2. Its final position is 6 m from its initial position at the end of that acceleration. For how much time did it accelerate?

PLEASE ANSWER!!!

Answer 1

here;

initial velocity, u = 0m/s

acceleration, a = 4m/s²

distance, s = 6m

we have to find time taken

so by using second equation of motion

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies \: 6 = 0 \times t + \frac{1}{2} \times 4 \times {t}^{2} \\ \\ \implies \: 6 = 2 {t}^{2} \\ \\ \implies \: {t}^{2} = \frac{6}{3} \\ \\ \implies \: {t}^{2} = 2 \\ \\ \implies \: t = \sqrt{2} \\ \\ \implies \: t = 1.23(approximately)$

so time taken is 1 second approximately ;

i think there is such mistake in your question please make ensure:)

Answer 2

here;

• initial velocity, u = 0m/s

• acceleration, a = 4m/s²

• distance, s = 6m

we have to find time taken

so by using second equation of motion

$\begin{gathered}s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies \: 6 = 0 \times t + \frac{1}{2} \times 4 \times {t}^{2} \\ \\ \implies \: 6 = 2 {t}^{2} \\ \\ \implies \: {t}^{2} = \frac{6}{3} \\ \\ \implies \: {t}^{2} = 2 \\ \\ \implies \: t = \sqrt{2} \\ \\ \implies \: t=1.23(approximately)\end{gathered}$

so time taken is 1 second approximately ;

i think there is such mistake in your question please make ensure:)