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)`

Step-by-step 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 (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)`