Question

a car is moving 18.2 starts to coast UP a frictionless 10.0 hill. How far does it roll before coming to a stop?

Answer 1

Answer:

The car travels 98.4m before coming to the stop !

Explanation:

First let "$s$" be the distance up the plane.

Then, By conservation of energy.

$\frac{1}{2} m(18.3)^2 = mg$ $s$ $sin(10)$

$(18.3)^2 = 2g$ $s$ $sin(10)$

$s = \frac{(18.3)^2}{2gsin(10) }$

$= 98.39564526$

$= 98.4 m$

Answer 2

97.3 m/s

The up answer has the right answer for THEIR equation. but that is NOT helpful to you since you asked a different question. You asked for 18.2 not 18.3. so heres the right answer.