A bullet moving horizontally to the right (+x direction) with a speed of 500 m/s strikes a sandbag and penetrates a distance of 10.0 cm. What is the average acceleration, in m/s2, of the bullet?

To find the average acceleration of the bullet, we need to use the kinematic equation that relates initial velocity (v_i), final velocity (v_f), acceleration (a), and distance (d). Since the bullet is coming to a stop, v_f is 0 m/s. We are given:

- The initial velocity, v_i = 500 m/s (in the +x direction) - The final velocity, v_f = 0 m/s (since the bullet stops) - The distance penetrated, d = 10.0 cm = 0.1 m (we need to convert cm to m)

We'll use the following kinematic equation:

v_f^2 = v_i^2 + 2ad

Solving for acceleration (a):

a = (v_f^2 - v_i^2) / (2d)

We can plug in our values now to calculate the acceleration:

a = (0^2 - (500 m/s)^2) / (2 * 0.1 m) a = (-250000 m^2/s^2) / (0.2 m) a = -1250000 m/s^2

The acceleration is negative because it’s a deceleration - the bullet is slowing down.

So the average acceleration of the bullet while it's in the sandbag is -1,250,000 m/s².