College Physics 2.30 – Acceleration of a fireworks shell

Solution:

We are given the following values:
$\text{v}_0=0\:\text{m/s}$
$\text{v}=65\:\text{m/s}$
$\Delta \text{x}=0.250\:\text{m}$

Part B

Solve for the acceleration first using the formula

$\text{v}^2=\text{v}_0^2+2\text{a}\Delta \text{x}$

Substitute the given values

$\left(65.0\:\text{m/s}\right)^2=\left(0\:\text{m/s}\right)^2+2\text{a}\left(0.250\:\text{m}\right)$

$\displaystyle \text{a}=\frac{\left(65.0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)}$

$\text{a}=8450\:\text{m/s}^2$

Part A

Substitute this value of acceleration to the formula $\text{t}=\frac{\text{v}-\text{v}_0}{\text{a}}$  to solve for time.

$\displaystyle \text{t}=\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2}$

$\text{t}=7.6923\times 10^{-3}\:\text{s}$