College Physics by Openstax Chapter 2 Problem 30

A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.

Solution:

We are given the following values:v0=0 m/sv_0=0\:\text{m/s}; vf=65.0 m/sv_f=65.0\:\text{m/s}; and Δx=0.250 m\Delta x=0.250\:\text{m}.

We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.

Part B

Solve for the acceleration first using the formula

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

We solve for acceleration in terms of the other variables.

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

a=(vf)2−(v0)22Δxa=(65.0 m/s)2−(0 m/s)22(0.250 m)a=8450 m/s2  (Answer)\begin{align*} a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\ a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\ a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part A

To solve for the time of this motion, we shall use the formula

vf=v0+atv_f=v_0+at

Solving for time, tt, in terms of the other variables we have.

t=vf−v0at=\frac{v_f-v_0}{a}

We now substitute the values given, and the computed acceleration to find the time.

t=vf−v0at=65.0 m/s−0 m/s8450 m/s2t=7.6922 ×10−3 s  (Answer)\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\ t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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