Question

1. a bullet leaves a rifle with a muzzle velocity of 521 m/s. while accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. determine the acceleration of the bullet (assume a uniform acceleration). want? equation: rearrange: enter:

Explanation:

Step1: Identify knowns and unknowns

We know the initial velocity \( v_0 = 0 \, \text{m/s} \) (starts from rest), final velocity \( v = 521 \, \text{m/s} \), and displacement \( x - x_0 = 0.840 \, \text{m} \). We need to find acceleration \( a \). The kinematic equation that relates these quantities is \( v^2 = v_0^2 + 2a(x - x_0) \).

Step2: Rearrange the equation for \( a \)

Starting with \( v^2 = v_0^2 + 2a(x - x_0) \), subtract \( v_0^2 \) from both sides: \( v^2 - v_0^2 = 2a(x - x_0) \). Then divide both sides by \( 2(x - x_0) \) to solve for \( a \): \( a=\frac{v^2 - v_0^2}{2(x - x_0)} \).

Step3: Substitute the known values

Substitute \( v_0 = 0 \, \text{m/s} \), \( v = 521 \, \text{m/s} \), and \( x - x_0 = 0.840 \, \text{m} \) into the equation:
\( a=\frac{(521)^2 - 0^2}{2\times0.840} \)
First, calculate \( (521)^2 = 521\times521 = 271441 \). Then, calculate the denominator: \( 2\times0.840 = 1.68 \). Now, divide: \( a=\frac{271441}{1.68}\approx161572 \, \text{m/s}^2 \).

Answer:

The acceleration of the bullet is approximately \( \boldsymbol{1.62\times 10^{5} \, \text{m/s}^2} \) (or \( 161572 \, \text{m/s}^2 \) depending on significant figures).