A car travels east 2 km for 5 minutes

A car travels east 2 km for 5 minutes, then north 3 km for 8 minutes, and then west 4 km for 10 minutes. Determine the total distance traveled and the magnitude of displacement of the car. Also, what is the magnitude of the average velocity and the average speed?


Show me the final answer↓

Let us draw out the travel path of the car to better visualize the question like so:

A car travels east 2 km for 5 minutes

Note that our origin is at A


The total distance traveled by the car is:

Total distance = AB + BC + CD = 2 + 3 + 4 = 9 km = 9000 m


To find the displacement, we can write a position vector from A to D and calculate it’s magnitude.

r=\left\{(2-4)i+3j\right\} km

r_{magnitude}=\sqrt{(-2)^2+(3)^2}=3.6 km = 3600 m


To figure out the average velocity and speed, we need to first determine the length of time the car was in motion for.

Total time = 5 min + 8 min + 10 min = 23 min = 1380 s


The average velocity is displacement divided by time:

v_{avg}=\dfrac{displacement}{t}=\dfrac{3600}{1380}=2.61 m/s


The average speed is total distance divided by the total time of travel:

Average speed = \dfrac{distance}{time}=\dfrac{9000}{1380}=6.52 m/s


Final Answers:

Total distance = 9 km = 9000 m

Displacement = 3.6 km = 3600 m

Average velocity = 2.61 m/s

Average speed = 6.52 m/s


This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 12, question 12-78.

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