# Motion in One Dimesion: The tortoise and the Hare Race Kinematics

#### Palilth

A turtle and a rabbit engage in a footrace over a distance of 4.00km. The rabbit runs 0.500km and then stops for a 90min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a total time of 1.75h, the rabbit wins the race.

A) Calculate the avg speed of the rabbit.

~ Using the equation for avg speed: Change in Distance/ Change in time.

I divided 1.75h from 4.00km and got 2.29km/h as the average speed.

B) What was his average speed before he stopped for a nap? Assume no detours or turning back.

~ I know that he ran twice as fact the average speed after the nap so i multiplied 2.29km/h by 2 and got 4.58km/h.

Im not sure what to do next...

#### ChipB

PHF Helper
Part A is correct.

For part B, think in terms of total time the rabbit runs, which must equal the distance traveled in each leg divided by the velocity of that leg. If the velocity on the first leg is V_1, then the velocity of thesecond leg is 2xV_1. Given that he naps for 1.5 Hr, and the race has a total time of 1.75 Hr, you know the total time he actually runs is 0.25 Hr. So:

Distance of leg 1/speed of leg 1 + distance of leg 2/speed of leg 2 = total time

0.5 Km/V_1 + 3.5 Km/(2xV_1) = 0.25 Hr

Solve for V_1.

#### HallsofIvy

Yes, if he ran a total distance of 4 km in a total time of 1.75 hours then his average speed was 4/1.75= 2.29 km/h. That was the easy part!

Let v be the speed at which he runs the first 0.5 km in km/h. It takes him 0.5/v hours to run that part. He runs the remaining 3.5 km at twice that speed, 2v so that takes 3.5/(2v)= 1.75/v hours. Including the 90 minute= 1.5 hour nap, the total time was 0.5/v+ 1.75/v+ 1.5= 2.25/v+ 1.5= 1.75 h. From that 2.25/v= 1.75- 1.5= 0.25. 2.25= 0.25v so v= 2.25/0.25= 9 km/h.

As a check, at 9 km/h it would take 0.5/9= 0.056 h to run the first 0.5 km. At twice that speed, 18 km/h, it would take 3.5/18= 0.194 h to run the remaining 3.5 km. Including the 1.5 h nap, that would be a total time of 0.056+ 0.194+ 1.5= 1.75 hours as required.