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
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$ 
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
$S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m$
So, the bullet penetrates the sandbag 1.8 meters.