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 

Time vs Position
Time vs Position

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