Question

A bullet is fired through a board, 14.0 cm thick, with its line of motion perpendicular to the face of the board. if it enters with a speed of 450 m/s and emerges with a speed of 220 m/s, what is the bullet's acceleration as it passes through the board?

Answer 1

Answer:

$a = -5.66 \times 10^5 m/s^2$

Explanation:

Here we know

initial speed will be

$v_i = 450 m/s$

final speed is

$v_f = 220 m/s$

total distance moved through it is given as

$d = 14 cm$

now we will use kinematics to find the acceleration

$v_f^2 - v_i^2 = 2 a d$

$220^2 - 450^2 = 2(a)(0.14)$

$a = -5.66 \times 10^5 m/s^2$

Answer 2

Okay, here is my stab at this, I hope it helps!

You know the bullet's initial velocity, V₀ = 450 m/s
You know the final velocity, V = 220 m/s
You also know how long the bullet accelerates (actually decelerates), 14cm, or .14 m

With this information, you learn that you need this equation.

V² = V₀² + 2a (x - x₀), because we have all the information except a, which is the acceleration. So putting it into the equation, it looks like this.

(220m/s)² = (450m/s)² +2a(.14m - 0m)

I'll let you solve the rest, but here are some hints.  Your answer will be really big because the bullet slows down really quickly in a really small distance, and you answer will be negative, because this acceleration is causing the bullet to go slower, which is also called deceleration. Hope that helps!