Chapter 1, Problem 12P

(a)

To determine

The sketch $s$ versus $t$ for both tortoise and hare on the same graph.

Explanation of Solution

Given:

The distance to be covered is $1.00\text{km}$ .

The speed of tortoise is $2.00\text{m}/\text{s}$ .

The hare moves at a speed of $10.0\text{m}/\text{s}$ for $60.0\text{s}$ , rests for $10.0\text{min}$ and continues at a speed of $10.0\text{m}/\text{s}$ for $40.0\text{s}$ .

Formula Used:

Speed is the distance covered in unit time by an object.

Write the expression for distance covered by an object.

  $d=vt$ ...... (I)

Here, $d$ is the distance covered, $t$ is the required time $v$ is the speed of object.

Rearrange the above expression in terms of $t$ .

  $t=\frac{d}{v}$ ...... (II)

Calculation:

For tortoise:

Substitute $1.00\text{km}$ for $d$ and $2.00\text{m}/\text{s}$ for $v$ in equation (II).

  $\begin{array}{c}{t}_t=\frac{1.00\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)}{2.00\text{m}/\text{s}}\\ =500\text{s}\end{array}$

For hare:

Substitute $10.0\text{m}/\text{s}$ for $v$ and $60.0\text{s}$ for $t$ in equation (I).

  $\begin{array}{c}{d}_{h1}=\left(10.0\text{m}/\text{s}\right)\left(60.0\text{s}\right)\\ =600\text{ m}\end{array}$

Substitute $0\text{m}/\text{s}$ for $v$ and $10.0\text{min}$ for $t$ in equation (I).

  $\begin{array}{c}{d}_{h2}=\left(0\text{m}/\text{s}\right)\left(10.0\text{min}\right)\\ =0\text{ m}\end{array}$

Substitute $10.0\text{m}/\text{s}$ for $v$ and $40.0\text{s}$ for $t$ in equation (I).

  $\begin{array}{c}{d}_{h3}=\left(10.0\text{m}/\text{s}\right)\left(40.0\text{s}\right)\\ =400\text{ m}\end{array}$

The sketch $s$ versus $t$ for both tortoise and hare on the same graph is plotted below.

  

Here, $v_t$ is the velocity of tortoise and $v_h$ is the velocity of hare.

Conclusion:

The sketch $s$ versus $t$ for both tortoise and hare on the same graph is plotted below.

(b)

To determine

The animal who wins the race.

Answer to Problem 12P

Tortoise wins the race.

Explanation of Solution

Given:

The distance to be covered is $1.00\text{km}$ .

The speed of tortoise is $2.00\text{m}/\text{s}$ .

The hare moves at a speed of $10.0\text{m}/\text{s}$ for $60.0\text{s}$ , rests for $10.0\text{min}$ and continues at a speed of $10.0\text{m}/\text{s}$ for $40.0\text{s}$ .

Calculation:

For tortoise:

The time taken to cover $1.00\text{km}$ is $500\text{s}$ .

For hare:

The first time interval is $60.0\text{s}$ , second time interval is $600\text{s}$ and the final interval is $40.0\text{s}$ .

Total time to cover $1.00\text{km}$ is the sum of all interval of time.

  $\begin{array}{c}T=60.0\text{s}+600\text{s}+40.0\text{s}\\ =\text{700}\text{s}\end{array}$

Conclusion:

The tortoise takes less time than hare to cover the distance of $1.00\text{km}$ . Thus, tortoise wins the race.

(c)

To determine

The average speed of hare.

Explanation of Solution

Given:

The distance to be covered is $1.00\text{km}$ .

The speed of tortoise is $2.00\text{m}/\text{s}$ .

The hare moves at a speed of $10.0\text{m}/\text{s}$ for $60.0\text{s}$ , rests for $10.0\text{min}$ and continues at a speed of $10.0\text{m}/\text{s}$ for $40.0\text{s}$ .

Formula Used:

Write the expression for average speed.

  $v_{\text{avg}}=\frac{D}{T}$ ...... (III)

Here, $v_{\text{avg}}$ is the average speed, $D$ is the total distance travelled and $T$ is the total time.

Calculation:

Substitute $1000\text{m}$ for $D$ and $700\text{s}$ for $T$ in equation (III).

  $\begin{array}{c}{v}_{\text{avg}}=\frac{1000\text{m}}{700\text{s}}\\ =1.43\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the average speed of hare is $1.43\text{m}/\text{s}$ .