# Acceleration Power [closed]

Suppose we have a trolley currently at rest at point A, but the motor switches on, so that it accelerates and reaches point B, at point B, the velocity is 5m/s, Distance of A-B is 10m, mass of the trolley is 12kg, find the power required.

I attempt to work out like this:

Assume: a=acceleration, F=force acting on the trolley, m=mass, u=initial velocity, v=final velocity, s=distance traveled, t=total time needed to travel from A to B, P=power required

Therefore, P=F(s/t) =m(a)(s/t)

also, v^2-u^2=2(a)(s) a=(v^2-u^2)/(2s) =(5^2-0)/(2x10) =1.25

find t: v=u+at t=(v-u)/a =(5-0)/1.25 =4

Therefore, P=12(1.25)(10/4) =37.5W

so power is 37.5W, is this correct?

## closed as off-topic by John Rennie, Brandon Enright, Emilio Pisanty, Waffle's Crazy Peanut, DumpsterDoofusMar 27 '14 at 12:54

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No, it is not correct. Let $a$ be acceleration; $F$ force; $m$ mass; $v_0$ initial velocity; $v_f$ final velocity; $P$ power required; $x$ distance travelled and $t$ time taken. Hence, $$P=\frac{Fs}{t}=\frac{ma\ x}{t}$$ Then $a=\frac{v_f^2-v_0^2}{2x}=\frac{5^2-0^2}{2*10}=1.25\text{ m s}^{-2}$ and $t=\frac{2x}{v_0+v_f}=\frac{2*10}{0+5}=4\text{ s}$
Hence, power is given by: $$P=\frac{ma\ x}{t}=\frac{12*1.25*10}{4}=150\text{ W}$$
• It comes from $x=\frac{1}{2}(v_0+v_f)t$. – Ruben Mar 28 '14 at 6:07