Question

A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sandbag,how deep,in meters,did the bullet penetrate the sandbag

Answer 1

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is
`v_i=180~m/s`, the final velocity is
`v_i=0~m/s`, and the total time of the motion is
`\Delta t=0.02~s`, so the acceleration is given by

`a= (v_f-v_i)/(\Delta t) = -9000~m/s^2`
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using

`S=v_i t + (1)/(2)at^2 = 180~m/s \cdot 0.02~s + (1)/(2)(-9000~m/s^2)(0.02~s)^2=1.8~m`
So, the bullet penetrates the sandbag 1.8 meters.