Question

a bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. the bullet penetrates for a distance of 0.0621 m. determine the acceleration of the bullet while moving into the clay. (assume a uniform acceleration.)

Explanation:

Step1: Identify the kinematic equation

Use the kinematic equation \(v^{2}=v_{0}^{2}+2ax\). Here, the final velocity \(v = 0\) (bullet stops), initial velocity \(v_{0}=367\space m/s\), and displacement \(x = 0.0621\space m\).

Step2: Rearrange the equation for acceleration

Rearrange \(v^{2}=v_{0}^{2}+2ax\) to solve for \(a\). We get \(a=\frac{v^{2}-v_{0}^{2}}{2x}\).

Step3: Substitute the values

Substitute \(v = 0\), \(v_{0}=367\space m/s\), and \(x = 0.0621\space m\) into the formula:
\(a=\frac{0-(367)^{2}}{2\times0.0621}\)
First, calculate \((367)^{2}=367\times367 = 134689\)
Then, \(2\times0.0621=0.1242\)
So, \(a=\frac{- 134689}{0.1242}\approx - 1.08\times10^{6}\space m/s^{2}\)

Answer:

The acceleration of the bullet is approximately \(-1.08\times 10^{6}\space m/s^{2}\)