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Expert-verified Found in: Page 82 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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.

(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).

See the step by step solution

## Determination of speed of sprinter

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 $$10.8\;{\rm{m}}/{\rm{s}}$$.

## Determination of sketch of position and time for sprinter

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). ### Want to see more solutions like these? 