Question #4aa5a

2 Answers
Sep 4, 2015

Answer:

48 km/hr

Explanation:

Pick some distance the motorcyclist might have traveled (it doesn't matter what distance but for ease of calculation I chose 120 km).

Time spent traveling 120 km at 40 km/hr
#color(white)("XXXX")120" km" div 40 " km/hr" = 3 " hours"#

Time spent traveling 120 km at 60 km/hr
#color(white)("XXXX")120" km" div 60 " km/hr" = 2 " hours"#

Total time spent traveling #120 xx 2 = 240# km
#color(white)("XXXX")3 " hours" + 2 " hours" = 5 " hours"#

Average speed
#color(white)("XXXX")240" km" div 5" hours" = 48 " km/hr"#

Sep 4, 2015

Answer:

The answer is (b) #"48 km/h"#.

Explanation:

I want to provide an alternative approach to figuring out the average sspeed of the motorcyclist.

So, you know that the motorcyclist travels the same distance, let's say #d#, for both his trips.

Let's assume that it takes him #t_1# hours to travel the distance at #"40 km/h"#, and #t_2# hours to travel the same distance at #"60 km/h"#.

This means that you can write

#d = 40 * t_1" "# and #" "d = 60 * t_2#

This is equivalent to

#40 * t_1 = 60 * t_2 implies t_1 = 3/2 * t_2#

The average speed of the motorcyclist can be thought of as the total distance he covered divided by the total time needed.

Since he goes from #A# to #B# for one trip, and from #B# to #A# for the second, the total distance covered will be

#d_"total" = 2 * d#

The total time will be

#t_"total" = t_1 + t_2#

#t_"total" = 3/2 * t_2 + t_2 = 5/2 * t_2#

The average speed will thus be

#bar(v) = (2d)/(5/2 * t_2) = 4/5 * d/t_2#

But #d/t_2# is equal to the speed of his return trip, #"60 km/h"#. This means that you have

#bar(v) = 4/5 * 60 = color(green)("48 km/h")#