# Can someone help with this physics question?

A lion and a pig participate in a race over a 1.70 km long course. The lion travels at a speed of 17.0 m/s and the pig can do 3.30 m/s. The lion runs for 1.360 km and then stops to tease the slow-moving pig, which eventually passes by. The lion waits for a while after the pig passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.

(a)

How far (in m) is the pig from the finish line when the lion resumes the race?

After the faster animal resumes the race, they both travel for the same amount of time to the end of the race. How far does the faster animal travel during this time? If we represent the distance the slower animal has traveled by D, how far does the slower animal travel? m

(b)

For how long in time (in s) was the lion stationary?

We need to find the time between when the faster animal stops and then resumes the race.

We know the distance and speed the faster animal travels before stopping and hence can find the time t.

We know the distance and speed the slower animal travels before the faster animal resumes the race and hence can find the time t'. The time of interest is

Δt = t' − t.

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• A lion and a pig participate in a race over a 1.70 km long course. The lion travels at a speed of 17.0 m/s and the pig can do 3.30 m/s. The lion runs for 1.360 km and then stops to tease the slow-moving pig, which eventually passes by. The lion waits for a while after the pig passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.

(a)

How far d (in m) is the pig from the finish line when the lion resumes the race?

1700-1360 = 340 m

tℓ = 340/17 = 20.00 sec

d = Vp*tℓ = 3.3*20 = 66.0 m

(b)

For how long in time ts (in s) was the lion stationary?

ts = tp-tℓ = d/Vp-d/Vℓ = 1700*(1/3.3-1/17) = 415.15 sec

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• d = vt

t = d/v

When the pig passes the lion there are 340m left in the race.

The lion takes 340m / (17m/s) = 20s to run this distance.

The pig needs to be 20s from the finish when the lion resumes running if they are to finish at the same time.

d = 3.3m/s * 20s = 66m

The pig wlil be 66m from the finish line when the lion resumes the race.

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• The pig will run 3.3 / 17 of the distance the lion does in the same amount of time.

If the lion started 340 m from the finish and ended with the pig, then the pig must've been 340 * 3.3/17 m from the finish when the lion started.

The pig takes 515.15 s to complete the race at full speed so the lion ran for 515.15 s * 3.3 / 17 in total and stood still for the other 515.15 (1 - 3.3 / 17) s

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