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.

Solution:

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.

After reaching 6 cm distance the bullet stops so the final velocity will be zero. The question is to find the negative acceleration exerted by the bullet when it stops after covering 6 cm distance.

So the given quantities are

Initial Velocity u = 20 m/s

Hi

Final Velocity v = 0

Distance S = 6 cm = 0.06 m

Using Newton’s third equation of motion, we can easily determine the declaration:

(refer to the image).

ItzDopeGirl❣

Answer 2

Answer:

As the bullet hits a sand box with velocity 20m/s it becomes the initial velocity exerted by the bullet and thus it covers a distance of 6m in the sand box with this velocity .After reaching 6cm distance the bullet stops so the final velocity will be zero.