Agolfer 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?

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

i think c. but don't trust it.

step-by-step explanation:

rick would give her the best deal as he will charge $8.88 per hour