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,
[tex]v=u+at[/tex]Where a is the acceleration of the skier.
On substituting the known values,
[tex]\begin{gathered} 9.57=0+a\times1.97 \\ \Rightarrow a=\frac{9.57}{1.97} \\ =4.86\text{ m/s}^2 \end{gathered}[/tex]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,
[tex]s=ut+\frac{1}{2}at^2[/tex]On substituting the known values,
[tex]\begin{gathered} s=0+\frac{1}{2}\times4.86\times1.97^2 \\ =9.43\text{ m} \end{gathered}[/tex]The skier travels for 9.43 m in the given time.
