A bullet leaves the barrel at 853 m/s

Calculating the acceleration of a bullet

A bullet leaves the barrel of a Barrett M82 at approximately 853 m/s. If the bullet slows to a velocity of 520 m/s at a range of 1500 m, what is the acceleration of the bullet?

To find the acceleration of the bullet, we can use the formula:

Acceleration = change in velocity / change in time

Given data:

  • Initial velocity (u) = 853 m/s
  • Final velocity (v) = 520 m/s
  • Time taken (t) = 3.0 seconds

First, let's calculate the change in velocity:

Change in velocity = Final velocity - Initial velocity

Change in velocity = 520 m/s - 853 m/s = -333 m/s

Next, we can plug in the values into the formula to find the acceleration:

Acceleration = (-333 m/s) / (3.0 seconds)

Acceleration ≈ -111 m/s²

Therefore, the acceleration of the bullet is approximately -111 m/s². The negative sign indicates that the bullet is decelerating.

What is the acceleration of the bullet?

The acceleration of the bullet is approximately -111 m/s².

← Understanding cavitation parameter in pump operation Consider two electric bulbs with different resistances connected in parallel →

More Posts: