calculate the deceleration of a bullet initially travelling at 360 m/s, if it is brought to rest after travelling 8.5 cm into a

Question

11 months ago

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tree trunk which it has struck.

1 answer:

sp2606 [1] · Answer 1 · 2024-08-18T14:00:53+03:00

The de - acceleration of bullet from 360 m/s after it struck the tree trunk and was brought to rest is -152.54 x 10⁴ m/s².

<h3>What is Acceleration in Kinematics?</h3>

Acceleration is defined as the rate of change of velocity with respect to time. Mathematically -

Given is a bullet initially travelling at 360 m/s struck into a tree trunk 8.5 cm inside and comes to rest. From this, we can write

initial velocity [u] = 360 m/s

final velocity [v] = 0 m/s

distance travelled [d] = 8.5 cm = 0.085 m

The time span after which bullet comes to rest will be -

Δt = d/u

Δt = 0.085/360 = 2.36 x 10 ⁻⁴ seconds

a = Δv/Δt

Δv = 0 - 360 = - 360 m/s

a = (v - u)/Δt

a = - 360/(2.36 x 10 ⁻⁴)

a = - 152.54 x 10⁴ m/s²

Therefore, the de - acceleration of bullet from 360 m/s after it struck the tree trunk and was brought to rest is -152.54 x 10⁴ m/s².

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