Question

A skier (68kg) starts from rest but then begins to move downhill with a net force of 92 N for 8.2 s. The hell levels out for 3.5 seconds. On this part of the hill, the net force on the skier is 22 N backwards.
(a) calculate the speed of the skier after 8.2 s
(b) Calculate the speed of the skier at the end of the section where the hill levels out
(c) Calculate the total distance travelled by the skier before coming to rest

Answer 1

(a) 11.1 m/s

The acceleration of the skier is given by:

$a=\frac{F}{m}$

where F = 92 N is the net force and m = 68 kg. Substituting,

$a=\frac{92 N}{68 kg}=1.35 m/s^2$

After 8.2 s, the speed of the skier is

$v=at$

where

a = 1.35 m/s^2 is the acceleration

t = 8.2 s is the time

Substituting,

$v=(1.35 m/s^2)(8.2 s)=11.1 m/s$

(b) 10.0 m/s

In this section of the hill, the net force is F = -22 N backwards. So, the acceleration of the skier is

$a=\frac{F}{m}=\frac{-22 N}{68 kg}=-0.32 m/s^2$

When starting this section, the skier is moving at u = 11.1 m/s. So, the final speed will be:

$v=u+at$

And substituting t=3.5 s, we find

$v=11.1 m/s+(-0.32 m/s^2)(3.5 s)=10.0 m/s$

(c) 237.9 m

The distance travelled by the skier in the downhill section is

$d=\frac{1}{2}at^2$

where a = 1.35 m/s^2 and t = 8.2 s. Substituting,

$d=\frac{1}{2}(1.35 m/s^2)(8.2 s)^2=45.4 m$

The distance travelled by the skier in the levelled out section is given by

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

where

v = 0 is the final speed

u = 11.1 m/s is the initial speed

a = -0.32 m/s^2 is the acceleration

d is the distance

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-(11.1 m/s)^2}{2(-0.32 m/s^2)}=192.5 m$

So, the total distance is

d = 45.4 m + 192.5 m = 237.9 m