College Physics 2.20 – Acceleration of an Olympic-class sprinter


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 \displaystyle \text{a}=4.50\:\text{m/s}^2,\:\text{t}=2.40\:\sec ,\:\text{and}\:\text{v}_0=0\:\text{m/s}

Part A

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

\displaystyle \text{v}_{\text{f}}=\text{v}_0+\text{at}

Substituting the given values,

\displaystyle \text{v}_{\text{f}}=0\:\text{m/s}+\left(4.50\:\text{m/s}^2\right)\left(2.40\:\text{s}\right)=108\:\text{m/s}

Part B

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

\displaystyle \text{x}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2

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

\displaystyle \text{x}=0+\frac{1}{2}\left(4.50\:\text{m/s}^2\right)\left(\text{t}^2\right)=2.52\text{t}^2

The values of the position given the time is tabulated below

Time (s)Position (m)
00
12.25
29
320.25
436
556.25
681
7110.25
8144
9182.25
10225

The values are plotted in the coordinate axes 

Time vs Position