20 PE
Question
An Olympic-class sprinter starts a race with an acceleration of \({\bf{4}}{\bf{.50}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.
Step-by-Step Solution
Verified(a) The speed of sprinter is \({\bf{10}}{\bf{.8}}\;{\bf{m/s}}\).
(b) The sketch of position and time for sprinter is given in Figure (1).
Given Data:
The acceleration of sprinter is \(a = 4.50\;{\rm{m}}/{{\rm{s}}^2}\)
The initial speed of car is \(u = 0\;{\rm{m}}/{\rm{s}}\)
The duration for speed of sprinter is \(t = 2.40\;{\rm{s}}\)
The speed of sprinter is found by using the first equation of motion as the acceleration of sprinter is uniform.
The speed of sprinter is given as
\(v = u + at\)
Here, \(v\) is the speed of sprinter.
Substitute all the values in the above equation.
\(\begin{array}{c}v = 0\;{\rm{m}}/{\rm{s}} + \left( {4.50\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {2.40\;{\rm{s}}} \right)\\v = 10.8\;{\rm{m}}/{\rm{s}}\end{array}\)
Therefore, the speed of sprinter is \(10.8\;{\rm{m}}/{\rm{s}}\).
The position of sprinter is given as
\(X = ut + \frac{1}{2}a{t^2}\)
Here, \(X\) is the position of sprinter at time \(t\) .
Substitute all the values in the above equation.
\[\begin{array}{l}X = \left( 0 \right)t + \frac{1}{2}\left( {4.50\;{\rm{m}}/{{\rm{s}}^2}} \right){t^2}\\X = 2.25{t^2}\end{array}\]
Sketch the above equation for the position and time graph of the sprinter.
It is as given in Figure (1).