Question

A bullet hits a Sand box with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sand box.

Answer 1

Answer:

The deceleration is -3333.33m/s².

Explanation:

The deceleration is defined as the negative acceleration exerted by the body to come to rest from motion. As the bullet hits a sand box with velocity 20 m/s, it becomes the initial velocity exerted by the bullet and thus it covers a distance of 6 cm in the sand box with this velocity.

Given,

Initial Velocity, u = 20m/s,  final velocity, v = 0m/s and Distance, s = 6cm = 0.06m (Distance is always in meters, so convert cm to m.)

Using equations of motion, we can easily determine the deceleration:

v^2 = u^2 + 2as

a = v^2 - u^2 / 2s = 0 - (-20)^2 / 2*0.06

==> a = -400/0.12 = -3333.33m/s²

Thank you:))

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