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Q40PE

Expert-verifiedFound in: Page 83

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)** $12.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s,$}\right.4.07\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

**b)** $11.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

**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:

$100-d=S\times 6.69$ ........(1)

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

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

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

$\begin{array}{rcl}100-d& =& S\times 6.69\\ 100-1.5\times S& =& S\times 6.69\\ S& =& 12.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The maximum speed of Bolt is $12.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .

The acceleration of the body can be calculated as:

$a=\frac{V-u}{t}\phantom{\rule{0ex}{0ex}}a=\frac{12.21-0}{3}\phantom{\rule{0ex}{0ex}}a=4.07\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${s}^{2}$}\right.$

Hence the acceleration of Bolt is $4.07\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${s}^{2}$}\right.$

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:

$200-d"=S"\times 16.3$ ......(3)

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

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

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

$\begin{array}{rcl}200-d"& =& S"\times 16.3\\ 200-1.5\times S"& =& S"\times 16.3\\ S"& =& 11.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The maximum speed of Bolt is $11.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .

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