Question

On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the trunk. If the meteorite struck the car with a speed of 130 m/s, what was the magnitude of its deceleration, assuming it to be constant?

Answer 1

Answer: $38409.09 m/s^{2}$

Explanation:

Since in this situation we assume the acceleration is constant, we can use the following equation:

$V^{2}={V_{o}}^{2} +2ad$ (1)  

Where:  

$V=0$ is the meteorite's final velocity  

$V_{o}=130m/s$ is the meteorite's initial velocity  (just in the moment it struck the car)

$a$ is the constant acceleration

$d=22 cm \frac{1 m}{100 cm}=0.22 m$ is the meteorite's traveled distance after the strike

Rewritting (1):

$0={V_{o}}^{2} +2ad$ (2)  

Clearing $a$:

$a=\frac{-{V_{o}}^{2}}{2d}$ (3)  

$a=\frac{-{(130 m/s)}^{2}}{2(0.22 m)}$ (4)  

$a=-38409.09 m/s^{2}$ (5)   This is the acceleration of the meteorite, the negative sign indicates is directed downwards

However, its magnitude is always positive. Therefore the answer is $a=38409.09 m/s^{2}$