PRACTICE 2: Steve the Stegosaur in grazing in a wooded valley. Over the course of 2 hourshe walks 100m North, 50 m West, 75 m South, 20 m North, 45 m East, and 60 m North.(Hint!!! Remember that there are 60 seconds in one minute and 60 minutes in onehour!!!!)1. What was Steve's total distance in meters traveled at the end of the trip?2. What was Steve's total displacement in meters (include direction!!)? Draw the vectors!!!3. What was Steve's average Speed in m/s?4. What was Steve's average Velocity in m/s?

Answered: PRACTICE 2: Steve the Stegosaur in…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

PRACTICE 2: Steve the Stegosaur in grazing in a wooded valley. Over the course of 2 hours
he walks 100m North, 50 m West, 75 m South, 20 m North, 45 m East, and 60 m North.
(Hint!!! Remember that there are 60 seconds in one minute and 60 minutes in one
hour!!!!)
1. What was Steve's total distance in meters traveled at the end of the trip?
2. What was Steve's total displacement in meters (include direction!!)? Draw the vectors!!!
3. What was Steve's average Speed in m/s?
4. What was Steve's average Velocity in m/s?

Expert Solution

Step 1

$Solution : Let the direction be indicated as shown in the figure . Let steve be at position O (0, 0) at start then we have : \begin{array}{rcl} first he walks 100 m, North & = & 100 \hat{j} \\ he walks 50 m, West & = & - 50 \hat{i} \\ he walks 75 m, South & = & - 75 \hat{j} \\ he walks 20 m, North & = & 20 \hat{j} \\ he walks 45 m, East & = & 45 \hat{i} \\ he walks 60 m, North & = & 60 \hat{j} \end{array} Total distance travelled by steve will be : D = 100 m + 50 m + 75 m + 20 m + 45 m + 60 m = 350 m - 1 Thus, the final position of steve in x - y coordinate will be : 100 \hat{j} - 50 \hat{i} - 75 \hat{j} + 20 \hat{j} + 45 \hat{i} + 60 \hat{j} \Rightarrow - 5 \hat{i} + 105 \hat{j} We know that : Displacement = final position - initial position = - 5 \hat{i} + 105 \hat{j} - 0 \hat{i} - 0 \hat{j} = - 5 \hat{i} + 105 \hat{j} Magnitude of displacement will be : |d| = \sqrt{{(- 5)}^{2} + {(105)}^{2}} = 105.11 m$