Question

Where do I start?

A driver sets out on a journey. For the first half of the distance she drives at a leisurely pace of 30 mph, she drives the second half at 60 mph. what is her average speed on the trip?
Hint: This problem is not as easy as it appears.

Answer 1

I think I have to plug it into d=16t^2 when d is in ft and t is in sec

Answer 2

Pick an arbitrary distance, say 120 miles to make the arithmetic easy.

1st half of the distance = 120 / 2 = 60. 60 miles @ 30 mph = 2 hours
2nd half of the distance = 60 miles @ 60 mph = 1 hour.

So, 120 miles are traveled in 1+2 = 3 hours -> avg speed is 120 miles / 3 hours = 40 mph.

You can also do this without naming a specific distance. Let the total distance driven be 2D. D miles are driven at 30 mph, and take D/30 hours. D miles are driven at 60 mph, and take D/60 hours. Total miles driven / total time taken = average speed

Total time taken is \[\frac{D}{30} + \frac{D}{60} = \frac{2D}{60}+\frac{D}{60} = \frac{3D}{60}\]

Average speed = total miles/ total time = \[\frac{2D}{(\frac{3D}{60})}\]but dividing by a fraction is just multiplying by the reciprocal fraction so\[2D *\frac{60}{3D}=40\]after the Ds cancel out.

Answer 3

awesome thank you!

Answer 4

Do you see why the distance doesn't matter?

Answer 5

yes I do. I had 3 hours of calc tonight so my brain is mush.

Answer 6

Let's hope it solidifies in the right shape :-)