Answer :
Answer:
The tortoise won the race by approximately [tex]33\; {\rm s}[/tex].
Explanation:
Note that speed in the question is measured in meters-per-second. Apply unit conversion and ensure that times in this question are also measured in seconds:
[tex]\begin{aligned} &1\; {\text{hour}} + 15\; {\text{minute}} \\ =\; & 1\; {\text{hour}} \times \frac{3600\; {\rm s}}{1\; {\text{hour}}} + 15\; {\text{minute}} \times \frac{60\; {\rm s}}{1\; \text{minute}} \\ =\; & 4500\; {\rm s}\end{aligned}[/tex].
Time required for the tortoise to finish the race:
[tex]\begin{aligned} \text{time} &= \frac{\text{distance}}{\text{speed}} \\ &= \frac{1000\; {\rm m}}{0.20\; {\rm m\cdot s^{-1}}} \\ &= 5000\; {\rm s}\end{aligned}[/tex].
The time required for the hare to finish the race includes:
- time required to cover [tex]800\; {\rm m}[/tex] at a speed of [tex]2.0\; {\rm m\cdot s^{-1}}[/tex],
- [tex]1\; \text{hour}[/tex] and [tex]15\; \text{minute}[/tex] ([tex]4500\; {\rm s}[/tex]) of nap, and
- time required to cover [tex](1000 - 800)\; {\rm m} = 200\; {\rm m}[/tex] at [tex]1.5\; {\rm m\cdot s^{-1}}[/tex].
[tex]\begin{aligned} \frac{800\; {\rm m}}{2.0\; {\rm m\cdot s^{-1}}} + 4500\; {\rm s} + \frac{200\; {\rm m}}{1.5\; {\rm m\cdot s^{-1}}} \approx 5033\; {\rm s} \end{aligned}[/tex].
Therefore, the tortoise wins the race by approximately [tex]5033\; {\rm s} - 5000\; {\rm s} = 33\; {\rm s}[/tex].