Question

Please help me with the following problem:

Joe is on a cross-country trip from New York to Los Angeles. He drives for 45.0 min at 70.0 km/h, 10.0 min at 95 km/h, and 30.0 min at 55.0 km/h, and he spends 20.0 min eating lunch and buying gas. What is the average speed for this part of his trip?
a. 67.6 km/h

b. 68.4 km/h - I did b as a guess, because I couldn't figure the answer, but I got it wrong

c. 54.7 km/h

d. 53.8 km/h

Two things that I did to help me were:

Delta d
--------
Delta t

and

m - m
-------
s - s

distance is m - m is meters
time is s - seconds

For the second one I did kilometer over hour.

I have no idea on how to do this, and I have no idea of if I am doing it right or not. I have tried distance over time tons of times, and I get different answers than the ones above. Can you guys help me please? I will greatly appreciate it.

Below is what bobpursley did to try to help me. I have not tried his way, and I do not know how to do his way. With his way, the lesson did not teach me how to do this.

bobpursley said this:

"average velocity=distance/time

distance=3/4*70+1/6*95 + 1/2*55
time= (45+10+30)/60

check my thinking. I get none of the answers. "

I am confused too, because he turn the minuted into seconds, but in fraction form. My home-school lesson did not teach me that. Please help me, I have been stuck on this problem for a couple days!

Answer 1

a. D = d1 + d2 + d3 =
70*(45/60) + 95*(10/60) + 55*(30/60) =
95.83 km.

T = 45 + 10 + 30 + 20=105 min=1.75 h.

Avg. Speed=D/T=95.83km/1.75h=54.8 km/h.


NOTE: My procedure is the same as Bob's,
but he forgot to include the 20 min for
eating and buying gas. Your answer in
Part a does not include the 20 min either.

Answer 2

Thank you Henry! :)

Answer 3

This exact queston is on my edenuity...thank you