Hi.

I have a little problem with this question. I did the first two (a) and (b), hopefully I did it ok.

I don't know how to do (c)... if anyone can help I'll be thankful.:)

Printable View

- Mar 3rd 2007, 01:30 AMPhoebe83Physics homework... question
Hi.

I have a little problem with this question. I did the first two (a) and (b), hopefully I did it ok.

I don't know how to do (c)... if anyone can help I'll be thankful.:) - Mar 3rd 2007, 03:00 AMticbol
A nice exercise on Distance = rate*time

Or distance = (average speed)(time)

-------------------------

a) Half the time at 35 mph, and half the time at 55 mph.

Let D = distance from San Antonio to Houston, in miles.

And T = total time of travel from San Antonio to Houston.

distance = rate*time

35(T/2) +55(T/2) = D

Multiply both sides by 2,

35T +55T = 2D

90T = 2D

Since average speed is D/T, then,

Divide both sides by T,

90 = 2D/T

D/T = 90/2 = 45 mph --------------answer.

------------------------------

b) Half the distance at 35 mph, and half the distance at 55 mph.

time = distance/rate

(D/2)/35 +(D/2)/55 = T

D/70 +D/110 = T

Multiply both sides by 70*110,

110D +70D = 70*110*T

180D = 7700T

D/T = 7700/180 = 42.7777...mph, average speed -------answer.

--------------------------

c) The average speed for the whole trip.

total time = distance/(ave. speed)

a) T1 = D/45 hrs

b) T2 = D/42.7777.... hrs

So, average speed for a) and b) is

= 2D/(T1 +T2)

= 2D/(D/45 +D/42.7777...)

= 2[(45)(42.7777..)] / [42.7777... +45]

= 43.86 mph ----------------------------answer.