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?

Solution:

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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