91Ó°ÊÓ

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).