Distance, rate, and time

Question

Alyssa left downtown Houston, and three hours later, Joan left going 54 mph faster to catch up. After another two hours, Joan caught up. Find Alyssa's average speed.

Answer 1

No, Neera, I do not agree.

The problem asks for Alyssa's average speed; you just ASSUMED it!

What the problem says is that Joan traveled 54 mph faster than Alyssa.

What is known is that Alyssa traveled 5 hours at some average speed and met Joan, who traveled 2 hours caught up with Alyssa; i.e. they both traveled the same distance.

Therefore for Alyssa: distance traveled=5r and Joan traveled 2(r+54)

5r=2(r+54) so that (Alyssa's rate is 36 mph and Joan's is 90...and they both traveled 180 miles.

Answer 2

Alyssa left 3 hours before with speed 54 mph and Joan met her after 2 hours -

Suppose Alyssa covered distance (d) in 5 hours with the speed of 54mph.

Alyssa covers in 1 hour = 54 miles

She will cover in 5 hours d = 54 x 5 = 270 miles

Joan has to cover 270 miles in 2 hours to meet Alyssa.

His speed should be = distance to cover / no of hours

= 270 / 2

= 135 mph

Answer : 135 mph