Question

When a gun is fired, a bullet accelerates through the gun barrel, a distance of .855 m, and exits the barrel with a speed of 457 m/s. What is the magnitude of the acceleration of the bullet?

Answer 1

Answer on Question #43405 – Physics - Mechanics | Kinematics | Dynamics

When a gun is fired, a bullet accelerates through the gun barrel, a distance of .855 m, and exits the barrel with a speed of 457 m/s. What is the magnitude of the acceleration of the bullet?

Solution:

$$d = 0.855\,\mathrm{m} - \text{traveled distance};$$ $$V = 457\,\frac{\mathrm{m}}{\mathrm{s}} - \text{final speed of the bullet};$$ $$t - \text{time of traveling};$$ $$a - \text{acceleration of the bullet};$$

Rate equation for the bullet:

$$V = a \cdot t$$ $$t = \frac{V}{a}$$

Equation of motion for the bullet ($V_0 = 0 - \text{initial velocity of the bullet}$)

$$d = \frac{a t^2}{2}$$

(1) in (2):

$$d = \frac{a \left(\frac{V}{a}\right)}{2} = \frac{V^2}{2a}$$ $$a = \frac{V^2}{2d} = \frac{\left(457\,\frac{\mathrm{m}}{\mathrm{s}}\right)^2}{2 \cdot 0.855\,\mathrm{m}} = 122\,100\,\frac{\mathrm{m}}{\mathrm{s}^2}$$

Answer: magnitude of the acceleration of the bullet is equal to $122\,100\,\frac{\mathrm{m}}{\mathrm{s}^2}$.