a bullet initially moving with a velocity 20m/s strikes a target and comes to rest after penetrating a distance 10cm in
Surely there is a question here.
the target calculate the deceleration caused by the target
it is its continuation
To solve this problem, we need to apply the principles of motion and use the equation of motion to find the deceleration of the bullet.
Let's break down the problem into known values:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s (as the bullet comes to rest)
Distance traveled (s) = 10 cm = 0.1 m
Acceleration (a) = ?
We can use the equation of motion:
v^2 = u^2 + 2as
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation:
a = (0^2 - 20^2) / (2 * 0.1)
Simplifying the equation:
a = (-400) / (0.2)
a = -2000 m/s²
The negative sign indicates that the bullet is decelerating, meaning it is being slowed down.
Therefore, the deceleration of the bullet is -2000 m/s².