What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 8.0 m/s when going down a slope for 5.0s and how far does the skier travel in this time?

1 Answer
May 25, 2017

Acceleration: 1.6 "m"/("s"^2)1.6ms2

Distance: 20. "m"20.m

Explanation:

We can use the equation

v_x = v_(0x) + a_xtvx=v0x+axt

To find the acceleration, assuming it's constant, of this motion. The initial velocity v_(0x)v0x is 00, as it started from rest. Thus, the acceleration is

a_x = (v_(x))/t = (8.0 "m"/"s")/(5"s") = color(red)(1.6 "m"/("s"^2)ax=vxt=8.0ms5s=1.6ms2

Now, to find the distance Deltax the skier travels, we'll use the equation

Deltax = v_(0x)t + 1/2a_xt^2

Plugging in known variables, the distance Deltax the skier traveled is

Deltax = (0)(5.0 "s") + 1/2(1.6"m"/("s"^2))(5.0"s")^2 = color(blue)(20. "m"