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Found in: Page 83

### College Physics (Urone)

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

# (a) A world record was set for the men’s 100 m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of . If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200 m dash with a time of 19.30 s Using the same assumptions as for the 100 m dash, what was his maximum speed for this race?

a) $12.2m}{s,}4.07m}{s}$

b) $11.2m}{s}$

See the step by step solution

## Step 1: Maximum speed and acceleration

The relation between the distance d, time t, and the speed S is:

${\mathbf{S}}{\mathbf{=}}\frac{\mathbf{d}}{\mathbf{t}}$

Here S is the speed of the body, and t is the time.

a) The bolt travels a portion of the total distance with a constant acceleration and the remainder at a maximal and constant speed.

The bolt travels a total distance 100 m . If a bolt travels d distance while accelerating, the remaining distance is 100-d , which is the distance he traveled at maximum speed.

Bolt accelerates for 3.00 s and then travels at a constant speed for the remaining 6.69 seconds.

Substitute 100-d for d’, S for average speed, and 6.69 s for t in the above equation:

$100-d=S×6.69$ ........(1)

The distance traveled by Bolt while accelerating can be calculated as:

$\begin{array}{rcl}\mathrm{Distance}\mathrm{traveled}& =& \mathrm{Sped}×\mathrm{Time}\\ \mathrm{d}& =& \frac{\mathrm{S}+0}{2}×3\\ \mathrm{d}& =& 1.5×\mathrm{S}\end{array}$ ........(2)

Substituting the value of equation (2) in (1), we get:

$\begin{array}{rcl}100-d& =& S×6.69\\ 100-1.5×S& =& S×6.69\\ S& =& 12.2m}{s}\end{array}$

The maximum speed of Bolt is $12.2m}{s}$ .

The acceleration of the body can be calculated as:

$a=\frac{V-u}{t}\phantom{\rule{0ex}{0ex}}a=\frac{12.21-0}{3}\phantom{\rule{0ex}{0ex}}a=4.07m}{{s}^{2}}$

Hence the acceleration of Bolt is $4.07m}{{s}^{2}}$

## Step 2: Maximum speed when he travels 200 m

b)

The entire distance traveled by Bolt is 200 m similar to the portion (a).

If Bolt travels d distance while accelerating, the remaining distance is 200-d which he traveled at maximum speed.

Bolt accelerates for 3.00 s , then travels at a constant speed for the following 16.30 s .

Substitute 100-d for d’’, S’’ for average speed and 16.30 sfor t in the above equation:

$200-d"=S"×16.3$ ......(3)

The distance traveled by Bolt while accelerating can be written as:

$Dis\mathrm{tan}cetraveled=AvarageSpeed×Time\phantom{\rule{0ex}{0ex}}d"=\frac{S"+0}{2}×3\phantom{\rule{0ex}{0ex}}d"=1.5×S"$...... (4)

Substituting the value of equation (4) in (3), we get:

$\begin{array}{rcl}200-d"& =& S"×16.3\\ 200-1.5×S"& =& S"×16.3\\ S"& =& 11.2m}{s}\end{array}$

The maximum speed of Bolt is $11.2m}{s}$ .