General Physics Problem (Velocitys)

[denverhockeyfan]

. Homework Statement
A hare and a tortoise compete in a race over a course of 1.00km long. The tortoise crawls straight and steadily at its max speed of 0.200 m/s toward the finish line. the hare runs at its max speed of 8.00 m/s toward the goal for .800km and then stops to taunt the tortoise. how close to the goal can the hare let the tortoise approach before resuming the race, which the tortoise win in a photo finish? assume that, when moving, both animals move steadily at there respective max speeds.

3. The Attempt at a Solution
I kinda feel stupid. I know it takes 5,000 seconds for the tortoise to go 1000m, 4,000 to go 800m, and 1,000 to go the last 200m.
It takes 1,250 seconds for the hare to go the whole 1000m, 1,000 seconds to go too the 800m mark, and 250 seconds to go the last 200m of the race.

Just not sure how to put this together.

 

[Dick]

It easier if you split it up and work from both ends at the same time. The hare went .8km and then stopped. Where is the tortoise then? If the race ended at a photo finish and the hare started again 200m from the finish line, how far did the tortoise run after the hare started again? The difference between those two tortoise positions is how far the tortoise ran while the hare was taunting. How long was it taunting?

 

[baba89]

I would approach this problem by first setting up a mathematical model to represent the motion of the hare and tortoise. We can use the equation v=d/t, where v is velocity, d is distance, and t is time.

For the tortoise, we know that its maximum speed is 0.200 m/s and it travels a distance of 1.00 km. Plugging these values into the equation, we get t = d/v = 1000m / 0.200 m/s = 5000 seconds. This is the total time it takes for the tortoise to complete the race.

For the hare, we know its maximum speed is 8.00 m/s and it travels a distance of 0.800 km before stopping to taunt the tortoise. So, t = 800m / 8.00 m/s = 100 seconds. This is the time it takes for the hare to reach the 800m mark.

Now, we need to determine how close the hare can let the tortoise approach before resuming the race. This can be calculated by subtracting the time it takes for the hare to reach the 800m mark from the total time it takes for the tortoise to complete the race. So, the hare can let the tortoise approach for 5000 seconds - 100 seconds = 4900 seconds.

To determine the distance that the tortoise can cover in 4900 seconds, we use the equation d = vt, where d is distance, v is velocity, and t is time. Plugging in the values, we get d = 0.200 m/s * 4900 seconds = 980 m. This means that the hare can let the tortoise approach up to 980 m before resuming the race.

In a photo finish, the hare and tortoise would cross the finish line at the same time. So, the hare would resume the race when the tortoise is 20 m away from the finish line (1000 m - 980 m = 20 m).

In conclusion, using mathematical modeling and equations, we can determine that the hare can let the tortoise approach up to 980 m before resuming the race, which would result in a photo finish.