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:v_0=0\:\text{m/s}; v_f=65.0\:\text{m/s}; and \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
\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x
We solve for acceleration in terms of the other variables.
a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}
Substitute the given values
\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
Solving for time, t, in terms of the other variables we have.
We now substitute the values given, and the computed acceleration to find the time.
\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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