Asked by Sandhya
A speedboat heads west at 115 km/h for 16.0 min. It then travels at 56.0° south of west at 87.0 km/h for 9.0 min.
(a) What is the average speed for the trip?
(b) What is the average velocity for the trip in magnitude and direction?
(a) What is the average speed for the trip?
(b) What is the average velocity for the trip in magnitude and direction?
Answers
Answered by
drwls
(a) average speed =
(total distance travelled)/(elapsed time)
= [115*(16/60) + 87*(9/60)]/(25/60)
(You can factor out the 1/60 to make the calculation easier)
= 104.9 km/h
(b) Add the two distances covered vectorially, taking their direction into account. Divide that distance vector by the elapsed time to get the average velocity. Use the component method. The magnitude is the square root of the sum of the squares of the two velocity components
(total distance travelled)/(elapsed time)
= [115*(16/60) + 87*(9/60)]/(25/60)
(You can factor out the 1/60 to make the calculation easier)
= 104.9 km/h
(b) Add the two distances covered vectorially, taking their direction into account. Divide that distance vector by the elapsed time to get the average velocity. Use the component method. The magnitude is the square root of the sum of the squares of the two velocity components
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