Question

(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 9.57 m/s when going down a slope for 1.97 s? (b) How far does the skier travel in this time?(a)Number_______ Units _________(b)Number_______ Units _________

Answer 1

Given,

The initial velocity of the skier, u=0 m/s

The final velocity of the skier, v=9.57 m/s

The time duration, t=.97 s

(a)

From the equation of motion,

`v=u+at`

Where a is the acceleration of the skier.

On substituting the known values,

`\begin{gathered} 9.57=0+a*1.97 \\ \Rightarrow a=(9.57)/(1.97) \\ =4.86\text{ m/s}^2 \end{gathered}`

Thus the magnitude of average acceleration of the skier is 4.86 m/s²

(b)

The distance traveled by the skier is given by one of the equations of motion as,

`s=ut+(1)/(2)at^2`

On substituting the known values,

`\begin{gathered} s=0+(1)/(2)*4.86*1.97^2 \\ =9.43\text{ m} \end{gathered}`

The skier travels for 9.43 m in the given time.