College Physics by Openstax Chapter 2 Problem 20

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 a‾=4.50 m/s2, Δt=2.40 sec⁡, and v0=0 m/s\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 vfv_f. The formula in solving for vfv_f is

vf=v0+atv_f=v_0+at

Substituting the given values,

vf=0 m/s+(4.50 m/s2)(2.40 s)vf=108 m/s  (Answer)\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=v0t+12at2x=v_0t+\frac{1}{2}at^2

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

x=0+12(4.50 m/s2)(t2)x=2.52t2\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 

Time vs Position: College Physics 2.20 - Acceleration of an Olympic-class Sprinter
Time vs Position
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