SOLUTION: An auto travels at the rate of 35-km/h, N for 6.0mins, then NE for 7.0mins and finally at 20-km/h, S for 8.0mins. what is the average velocity of the auto? Whta is the average spee

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Question 983054: An auto travels at the rate of 35-km/h, N for 6.0mins, then NE for 7.0mins and finally at 20-km/h, S for 8.0mins. what is the average velocity of the auto? Whta is the average speed of the auto?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
It is not clearly specified,
but it seems like after going N for 6.0 minutes at 35 km/h,
the auto continues NE for 7.0 minutes at the same speed.
That is what I will assume.
The whole trip looks like this:
The average velocity is a vector with magnitude and direction,
and can be found as the start-end displacement vector divided by the travel time.
The average speed is a scalar quantity (a number with units of km/h),
and can be found as the total distance covered along the trajectory divided by the travel time.
The distances traveled are as follows:
red%28%22N+%3A%22%29 %2835km%2F%2260+minutes%22%29%2A%286minutes%29=3.5km
green%28%22NE+%3A%22%29 %2835km%2F%2260+minutes%22%29%2A%287minutes%29=49%2F12km=4.083km(rounded)
blue%28%22S+%3A%22%29 %2820km%2F%2260+minutes%22%29%2A%288minutes%29=8%2F3km=2.667km(rounded).
Now the trips looks like this:
The approximate total distance traveled is
3.5km%2B4.083km%2B2.667km=10.250km .
The total travel time is
6minutes%2B7minutes%2B8minutes=21minutes .
The average speed of an auto that travels a total of 10.250km in 21minutes is
%2810.250km%2F%2221minutes%22%29%2860minutes%2F%221hour%22%29=29.29%22km+%2F+h%22=29%22km+%2F+h%22 (rounded).
The eastward displacement is the horizontal leg of an isosceles right triangle with a hypotenuse of 49%2F12km=4.083km(rounded).
It's measure in km is x and 2x%5E2=%2849%2F12%29%5E2=49%5E2%2F12%5E2 ,
so x%5E2=49%5E2%2F%282%2A12%5E2%29---> .
That is also the vertical leg of the same isosceles right triangle,
which is the northward displacement during the NE portion of the trip.
The total (approximate) northward displacement in km is
3.5%2B2.887-2.667=3.72 .
Now the trips looks like this:
The magnitude of the displacement (in km) is approximately
sqrt%283.72%5E2%2B2.887%5E2%29=4.709 .The
The approximate) magnitude of the average velocity is
%284.709km%2F%2221minutes%22%29%2860minutes%2F%221hour%22%29=13.45%22km+%2F+h%22=13.5%22km+%2F+h%22 (rounded),
and red%28theta%29 gives us the direction.
tan%28red%28theta%29%29=2.887%2F3.72=0.7761--->red%28theta%29=37.8%5Eo ,
so the direction is 37.8%5Eo East of North.