Question

A typical male sprinter can maintain his maximum acceleration for $\mathbf{2}.\mathbf{0}\mathbf{s}$, and his maximum speed is $\mathbf{10}\mathbf{m}\mathbf{/}\mathbf{s}$. After he reaches this maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first $\mathbf{2}.\mathbf{0}\mathbf{s}$ of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) $\mathbf{50}.\mathbf{0}\mathbf{m}$; (ii) $\mathbf{100}.\mathbf{0}\mathbf{m}$; (iii) $\mathbf{200}.\mathbf{0}\mathbf{m}$?

Short answer

a) the sprinter has covered a distance of $10\mathrm{m}$before he reaches his maximum speed and b) the magnitude of the average velocity at i) $50\mathrm{m}$, ii) $100\mathrm{m}$and iii) $200\mathrm{m}$ are $8.33\mathrm{m}/\mathrm{s}$, $9.09\mathrm{m}/\mathrm{s}$ and $9.52\mathrm{m}/\mathrm{s}$respectively.

Step-by-step solution

123

The given data can be listed below as,

• The maximum speed of the sprinter is$10\mathrm{m}/\mathrm{s}$.
• The sprinter can maintain his acceleration for$2.0\mathrm{s}$.
• The acceleration of the male sprinter is constant in the first$2.0\mathrm{s}$.
• The acceleration of the sprinter is 0 when the sprinter reaches his maximum speed.

23

This law states that a particular object will continue to be in rest or in uniform motion unless it is resisted by an external object.

The equation of the displacement gives the distance the sprinter has to run and the equation of velocity gives the velocity of the sprinter.

123

a) From Newton’s first law, the equation of displacement for the sprinter can be expressed as:

$\begin{array}{rl}lx& =u\times t+\frac{1}{2}at.t\\ x& =ut+\frac{1}{2}\left(v-u\right)t\end{array}$

Here, x is the distance travelled by the sprinter, u is the initial velocity of the sprinter that is 0, v is the maximum velocity of the sprinter that is$10\mathrm{m}/\mathrm{s}$and t is the time taken by the sprinter that is$2\mathrm{s}$.

Substituting the values in the above equation, we get-

$\begin{array}{rl}lx& =0+\frac{1}{2}\left(10\mathrm{m}/\mathrm{s}-0\right)\times 2\mathrm{s}\\ x& =10\mathrm{m}\end{array}$

Thus, the sprinter has covered a distance of$10\mathrm{m}$before he reaches his maximum speed.

b) As the sprinter has run about$10\mathrm{m}$in constant acceleration, then it starts a constant velocity.

i) The displacement reduces to$50\mathrm{m}-10\mathrm{m}=40\mathrm{m}$.

Hence, from the first law of Newton, the time the sprinter took to cover the distance of$40\mathrm{m}$, is expressed as:

$t=\frac{s}{v}$

Here, s is the reduced displacement of the sprinter and v is the constant velocity of the sprinter.

Substituting the values in the above equation, it can be expressed as:

$\begin{array}{r}ct=\frac{40\mathrm{m}}{10\mathrm{m}/\mathrm{s}}\\ =4\mathrm{s}\end{array}$

From the Newton’s first law, the magnitude of the average velocity can be expressed as:

$v=\frac{s}{t}$

Here, s is the distance needed to cover the sprinter which is$50\mathrm{m}$and t is the additional time taken after$2\mathrm{s}$to cover the distance.

Substituting the values in the above equation, we get-

$\begin{array}{r}cv=\frac{50\mathrm{m}}{2\mathrm{s}+4\mathrm{s}}\\ =8.33\mathrm{m}/\mathrm{s}\end{array}$

ii)The displacement reduces to$100\mathrm{m}-10\mathrm{m}=90\mathrm{m}$.

Hence, from the first law of Newton, the time the sprinter took to cover the distance of$90\mathrm{m}$, is expressed as:

$t=\frac{s}{v}$

Here, s is the reduced displacement of the sprinter and v is the constant velocity of the sprinter.

Substituting the values in the above equation, it can be expressed as:

$\begin{array}{r}ct=\frac{90\mathrm{m}}{10\mathrm{m}/\mathrm{s}}\\ =9\mathrm{s}\end{array}$

From the Newton’s first law, the magnitude of the average velocity can be expressed as:

$v=\frac{s}{t}$

Here, s is the distance needed to cover the sprinter which is$100\mathrm{m}$and t is the additional time taken after$2\mathrm{s}$to cover the distance.

Substituting the values in the above equation, we get-

$\begin{array}{r}cv=\frac{100\mathrm{m}}{2\mathrm{s}+9\mathrm{s}}\\ =9.09\mathrm{m}/\mathrm{s}\end{array}$

iii)The displacement reduces to$200\mathrm{m}-10\mathrm{m}=190\mathrm{m}$.

Hence, from the first law of Newton, the time the sprinter took to cover the distance of$190\mathrm{m}$, is expressed as:

$t=\frac{s}{v}$

Here, s is the reduced displacement of the sprinter and v is the constant velocity of the sprinter.

Substituting the values in the above equation, it can be expressed as:

$\begin{array}{r}ct=\frac{190\mathrm{m}}{10\mathrm{m}/\mathrm{s}}\\ =19\mathrm{s}\end{array}$

From the Newton’s first law, the magnitude of the average velocity can be expressed as:

$v=\frac{s}{t}$

Here, s is the distance needed to cover the sprinter which is$200\mathrm{m}$and t is the additional time taken after$2\mathrm{s}$to cover the distance.

Substituting the values in the above equation, we get-

$\begin{array}{r}cv=\frac{200\mathrm{m}}{2\mathrm{s}+19\mathrm{s}}\\ =9.52\mathrm{m}/\mathrm{s}\end{array}$

Thus, the magnitude of the average velocity at i) $50\mathrm{m}$, ii) $100\mathrm{m}$and iii) $200\mathrm{m}$ are $8.33\mathrm{m}/\mathrm{s}$, $9.09\mathrm{m}/\mathrm{s}$ and $9.52\mathrm{m}/\mathrm{s}$respectively.