A student drives the 100 mi trip back to campus after springbreak and travels wi

Question

A student drives the 100 mi trip back to campus after springbreak and travels with an average speed of 61 mi/h for 1 hour and30 minutes. (a). what distance was traveled during this time?-------m (b). Traffic gets heavier, and the last part of the trip takesanother half-hour. What was the average speed during this leg ofthe trip? ------mi/h (c). Find the average speed for the total trip.----mi/h A student drives the 100 mi trip back to campus after springbreak and travels with an average speed of 61 mi/h for 1 hour and30 minutes. (a). what distance was traveled during this time?-------m (b). Traffic gets heavier, and the last part of the trip takesanother half-hour. What was the average speed during this leg ofthe trip? ------mi/h (c). Find the average speed for the total trip.----mi/h

Explanation / Answer

The crucial formula for this question is x= v(average) *t Here, x = the distance traveled, v(average) the average speed inmiles/hour and t is the time in hours a) x= v(average)*t where: v(average)= 61 mi/h and t = 1.5 h so: x = 61*1.5= 91.5 miles. b) totale distance = 100 miles distance traveled = 91.5 miles So the distance to travel is 100-91.5 = 8.5 miles. This takes 30 minutes, t = 0.5 hours x= v(average)*t x/t = v(average) = 8.5/0.5 = 17 miles/ h c) v(average) of the total trip v(average part 1 ) * t1 + v(average part 2) * t2 = 61 miles/h * 1.5 + 17 miles/h * 0.5 h = 100 miles in 2hours This means that the average speed is 50miles/h. (Note that velocity is always in speed per time unit, such asmiles/h or meters per second) Good Luck further!

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