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Q2.8-65PE

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Found in: Page 35

### College Physics (Urone)

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

# A graph of v(t) is shown for a world-class track sprinter in a $${\bf{100}} - {\bf{m}}$$ race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at $$t = {\rm{ }}{\bf{5}}{\rm{ }}{\bf{s}}$$? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

(a) $$6 m/s$$

(b) $$12{\rm{ m/s}}$$

(c) The acceleration is approximately to the value $$3 m/{s^2}$$

(d) $$10.3 s$$

See the step by step solution

## Graph and its analysis

To analyses movement, motion graphs can be employed.

The mathematical approaches and graphical solutions for determining motion equations are the same.

The slope of a graph of displacement x vs. time t represents velocity v.

The inclination of a graph of velocity v vs. time t is called acceleration.

Average velocity, instantaneous velocity, and acceleration can all be calculated using graphs.

## Average velocity for the first 4 s

a) Here if we carefully look into the figure, in your text book lets take any two points on the line

$$\begin{array}{c}average\,velocity = \frac{{{v_1} + {v_2}}}{2}\\ = \frac{{0 + 12}}{2}\\ = 6\,m/s\end{array}$$

Hence the average velocity of the body up to 4 s is 6 m/s

## instantaneous velocity at t = 5 s

b)

Hence when the time is 5s the velocity is constantly 12 m/s.

## Average acceleration between 0 to 4 s

(c)

Here if we carefully look into the figure, in your text book let’s take any two points on the line

Coordinate $$\left( {0,0} \right)$$ and $$\left( {4,12} \right)$$

Slop of the line will be

$$\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(0,0)\,\,\,and\,\,\,(4,12)\\m = \frac{{12 - 0}}{{4 - 0}}\\m = 3\,m/{s^2}\end{array}$$

Hence the acceleration is approximately to the value$$3 m/{s^2}$$.

## Time for the race

(d) For the first $$4{\rm{ s}}$$

Distance travelled is equal to area under the curve

Distance = $$\frac{1}{2} \times 4 \times 12 = 24m$$

He can run the remaining distance by

$$\begin{array}{c}t = \frac{{distance}}{{velocity}} = \frac{{76}}{{12}}\\ = 6.3 s\end{array}$$

Total time of the sprint is$${t_{total}} = 4 + 6.3 = 10.3 s$$