Physics 20—Module 1

Lesson 3—Graphical Analysis of Uniform Motion

 

Self-Check Answers

 

SC 12.

 

Given
 

three position-time graphs

 

Required

 

to find the velocity of each object from the graphs

 

Analysis and Solution

 

Find the slope of the velocity-time graph for the time interval given. The slope will be equal to the velocity of the object in that interval of time.

 

(a) The positive direction of the position is forward.

 



(b) The positive direction of the position is forward. The time of 10 minutes will be converted to seconds using a ratio set up, so the minutes units cancel.




(c) The positive direction of the position is forward.


 

Paraphrase

 

The velocity in each case is

 

(a) 1.0 m/s [backward]

(b) 0.033 m/s [left]

(c) 1.7 m/s [forward]

 

SC 13.

 

Given

 

       Δt =15.0 min

 

Required

 

the vehicle’s displacement ()

 

Analysis and Solution

 

The solution can be found algebraically by rearranging the average velocity formula. The time of 15.0 minutes will be converted to seconds using a ratio set up so that the minutes units cancel.

 

 

The positive value means the direction is west.

 

Paraphrase

 

The vehicle’s displacement is 27.0 km [W].

 

SC 14.

 

Given

 

position-time graph

 

Required

 

the average speed (vave), average velocity (), and net displacement ()

 

Analysis and Solution

 

The positive direction is toward the right.

 

dtotal = (25.0 m – 0.0 m) + |(-25.0 m) - (+25.0 m)|
           = 25.0 m + 50.0 m

          = 75.0 m

 

 

        = (–25.0 m) – (0.0 m)

        = –25.0 m             The negative value indicates the direction is left.

 

 

The negative value indicates the direction is left.

 

Paraphrase

 

The average speed is 3.75 m/s. The average velocity is 1.25 m/s [left], and the net displacement is 25.0 m [left].

 

SC 15.

 

Given

 

velocity-time graph

 

Required

 

the distance the elk will travel (Δd) in 30 min

 

Analysis and Solution

 

The area between a velocity-time graph and the time axis for a given interval of time is equal to the displacement of the object in that interval of time. Find the area under the graph line and the value without the direction will equal the distance. The time of 30 min = 0.50 h.

 

A = l x w

   = (35 km/h)(0.50 h)

   = 18 km, corrected to 2 significant digits

 

Paraphrase

 

The elk will travel 18 km.

 

Physics 20 © 2008 Alberta Education