College Physics 2.30 – Acceleration of a fireworks shell


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:
 \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}