Question

A car traveling at 30 m/s runs out of gas while traveling up a 10 slope. how far up the hill what coast before starting to roll back down?

Answer 1

Answer:

264.7 m

Explanation:

The net force felt by the car travelling along the slope is equal to the component of the weight parallel to the slope, so:

$F=-mg sin \theta$

where

m is the mass of the car

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=10^{\circ}$

and the negative sign is due to the fact the force is opposite direction to the motion of the car

The acceleration of the car is therefore:

$a=\frac{F}{m}=-gsin \theta=-9.8 sin 10^{\circ}=-1.7 m/s^2$

Now we can find how far the car went up the hill by using the equation:

$v^2-u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 30 m/s is the initial velocity

g = -1.7 m/s^2 is the acceleration

d is the distance covered

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-30^2}{2(-1.7)}=264.7 m$