# College Physics Problem 2.20

#### An Olympic-class sprinter starts a race with an acceleration of $4.50\:m/s^2$. (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

$a=4.50\:m/s^2$
$t=2.40\:s$
$v_0=0\:m/s$

a) The unknown is $v_f$. The formula in solving for $v_f$ is

$v_f=v_o+at$
$v_f=0\:m/s+\left(4.50\:m/s^2\right)\left(2.40\:s\right)$
$v_f=10.8\:m/s$

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

$x=v_ot+1/2at^2$
$x=0+1/2\left(4.50\:m/s^2\right)\left(t^2\right)$
$x=2.25t^2$

The values of the position given the time is tabulated below

Time (s) Position (m)
0 0
1 2.25
2 9
3 20.25
4 36
5 56.25
6 81
7 110.25
8 144
9 182.25
10 225

The values are plotted in the coordinate axes

You can now buy the complete solution manual for College Physics. Just complete the google form below.