Question

A bullet is moving at a speed of 40 m/s when it embeds into a lumps on a sponge wall. The bullet penetrates with the acceleration of 100m/s2 and then it stops. How much distance travelled by the bullet. (assume a uniform acceleration.)

Answer 1

Here, initial velocity of bullet is u=367m/s and final velocity is v=0m/s.

Distance traveled by bullet is d=0.0621m

If a be the acceleration of bullet.

Using v

2

−u

2

=2ad

0

2

−(367)

2

=2a(0.0621)

⟹ a=−1.08×10

−6

m/s

2

(negative sign indicates deceleration of bullet means it slows down).

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Answer 2

Answer:

Distance = 8m

Explanation:

Initial velocity = 40m/s

acceleration = -100m/s2

Final velocity = 0m/s

By using the equation 2aS = v^2  - u^2

Substitute the value

2 x -100 x S = 0 - (40)^2

200S = 0 - 1600

S = -1600 / -200

S= 8 m

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