A golfer rides in a golf cart at an average speed of 3.10 m/s for 28.0 s. She then gets out of the cart and starts walking at an average speed of 1.30 m/s. For how long (in seconds) must she walk: if her average speed for the entire trip, riding and walking, is 1.80 m/s?

Question

Mathematics

Mckenna Alvarado

2 April, 23:28

A golfer rides in a golf cart at an average speed of 3.10 m/s for 28.0 s. She then gets out of the cart and starts walking at an average speed of 1.30 m/s. For how long (in seconds) must she walk: if her average speed for the entire trip, riding and walking, is 1.80 m/s?

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Dixon · Answer 1

Answer: she must walk for 72.8 s

Hi!

Lets say that with the cart she rides a time T1 (28 s) for a distance D1, then the average speed in the cart is V1 = D1 / T1 = 3.10 m/s. We can calculate D1 = (28 s) * (3.10 m/s) = 86.8 m

She then walks a time T2 for a distance D2, with average speed

V2 = D2 / T2 = 1.30 m/s

For the entire trip, we have average speed:

V3 = (D1 + D2) / (T1 + T2) = 1.80 m/s

We can solve for T2:

(1.8 m/s) * (28s + T2) = 86.8 m + D2 = 86.8 m + (1.3 ms) * T2

Doing the algebra we get: T2 = 72,8 m/s