Question

a bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 5.00 × 10⁵ m/s² for 6.90 × 10⁻⁴ s. what is its muzzle velocity (in m/s) (that is, its final velocity)? (enter the magnitude.)

Explanation:

Step1: Recall the kinematic equation

The kinematic equation for final velocity \( v \) when initial velocity \( u = 0 \) (starts from rest), acceleration \( a \), and time \( t \) is \( v=u + at \). Since the bullet starts from rest, \( u = 0 \), so the equation simplifies to \( v=at \).

Step2: Substitute the given values

We are given \( a = 5.00\times 10^{5}\space m/s^{2} \) and \( t=6.90\times 10^{- 4}\space s \). Substitute these values into the equation \( v = at \):

\( v=(5.00\times 10^{5}\space m/s^{2})\times(6.90\times 10^{-4}\space s) \)

First, multiply the coefficients: \( 5.00\times6.90 = 34.5 \)

Then, multiply the powers of 10: \( 10^{5}\times10^{-4}=10^{5 - 4}=10^{1} \)

So, \( v = 34.5\times10^{1}\space m/s=345\space m/s \)

Answer:

\( 345 \)