Problem 5. A bullet leaves a rifle with a velocity of 452 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.93 m. Determine the acceleration of the bullet.

Answer:

1.1 x 10⁵m/s²

Explanation:

Given parameters:

Velocity  = 452m/s

distance  = 0.93m

Unknown:

Acceleration of the bullet  = ?

Solution:

To solve this problem, we use one of the kinematics equation which is given below:

           V² = U²  + 2aS

V is the final velocity

U is the initial velocity  = 0m/s

a is the unknown acceleration

S is the distance traveled  

 So;

          452²   = 0²  + (2 x a x 0.93)

          204304  = 1.86a

               a  = 1.1 x 10⁵m/s²

Final answer:

The acceleration of the bullet in the gun barrel can be calculated using the kinematic equation for motion. By substituying the given values into the equation, we find the acceleration to be approximately 1.095 x 10^5 m/s^2.

Explanation:

The subject of this question is Physics, specifically a topic under mechanics known as kinematics. The problem given can be solved using kinematic equations which are used to describe the motion of an object without considering the forces that cause it to move. In this case, the final velocity (vf) of the bullet is given as 452 m/s, the initial velocity (vi) is assumed to be 0 as it starts from rest, and the distance (d) is given as 0.93 m. We are asked to determine the value of acceleration (a).

Using the kinematic equation vf2 = vi2 + 2ad and substituting the given values, we get (452 m/s)2 = 0 + 2*a*0.93 m. We can rearrange to solve for acceleration to get: a = (452 m/s)2 / (2*0.93 m) = 109523.66 m/s2.

So, the acceleration of the bullet in the gun barrel is approximately 1.095 x 105 m/s2.

Learn more about Kinematics here:

brainly.com/question/35140938

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