Question

A bullet accelerates at a rate of 90,000m/s2 from rest. Calculate its velocity if it travels a distance of 0.5m while accelerating.

Answer 1

Explanation:

apply third law of motion

v^2=2*90000*.5

v=√90000

v=300

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Answer 2

The speed of the bullet if it travels a distance of 0.5 m while accelerating is 300 m/s.

Explanation:

Given that,

Acceleration of the bullet, $a=90000\ m/s^2$

Initial velocity of the bullet, u = 0

Let v is the velocity of the bullet if it travels a distance of 0.5 m while accelerating. Using third equation of motion :

$v^2-u^2=2ad\\\\v^2=2ad\\\\v^2=2\times 90000\times 0.5\\\\v=300\ m/s$

So, the speed of the bullet if it travels a distance of 0.5 m while accelerating is 300 m/s.

Learn more,

Equation of kinematics

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