# You drive  on Interstate   10 from  San Antonio  to Houston,  one- half  the time at 35.0 mi/h  (= 56.3  km/h)  and the other  half at 55.0 mi/h  (= 88.5 km/h). On the way back you travel one-half the distance  at 35.0 mi/h  and the other half at 55.0 mi/h. What is  your  average  speed  (a)  from  San Antonio   to  Houston,   (b) from  Houston  back  to San Antonio,  and (c) for the entire  trip?

8 years ago

Aparna Venkateshwaran
26 Points
6 years ago
Hi, I was solving this problem and I hit a snag in part c, where you say s is half the distance of s`. This implies s`=2s and not the other way around - won`t this change the answer for part c, because now a factor of 2 doesn`t cancel in one half of the denominator?
Rishi Sharma
2 years ago
Dear Student,

According to the problem while going to Houston the the half of the time the speed was 55 km/h and next half of the time was 90 km/h
While coming back on the first half the speed was 55 km/h and on the next half the speed was 90 km/h
a) Therefore the average speed is given by
S(avg) = total distance / t
Now total distance, s = u1x t/2 +u2 x t/2
where u1 = 55 km/hr and u2 = 90 km/hr
Therefore S(avg) = u1x t/2 +u2 x t/2 /t
= (t/2)(u1+u2)/t
= u1+u2/2
= 55+90/2 = 72.5 km/h
b) Therefore the average speed is given by
S(avg) = total distance/t
Now total distance is P
S(avg) = P/t1+t2
where t1 is the time taken for the displacement at u1 speed and t2 is for u2
we can write ,
t1 = p1/u1 = P/2/v1 and t2 = p2/u2 = P/2/u2
Therefore S(avg) = P/t1+t2
= P/P/2/u1+ P/2/u2
= 1/2x 55 + 2 x 90
= 68.27 km/h

Thanks and Regards