An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s².

(a) What is her speed 2.40 s later?

(b) Sketch a graph of her position vs. time for this period.

Solution:

We are given \overline{a}=4.50\:\text{m/s}^2, \ \Delta t=2.40\:\sec ,\:\text{and}\: v_0=0\:\text{m/s}

Part A

The unknown is v_f. The formula in solving for v_f is

v_f=v_0+at

Substituting the given values,

\begin{align*}
v_f & =0\:\text{m/s}+\left(4.50\:\text{m/s}^2\right)\left(2.40\:\text{s}\right) \\
v_f & = 108\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The relationship between position and time can be calculated using the formula

x=v_0t+\frac{1}{2}at^2

Then, with the given, we can express position in terms of time

\begin{align*}
x & =0+\frac{1}{2}\left(4.50\:\text{m/s}^2\right)\left(\text{t}^2\right) \\
x & =2.52\text{t}^2 \\
\end{align*}

The values of the position given the time are tabulated below

[wpdatatable id=2]

The values are plotted in the coordinate axes