SC 12.
three position-time graphs
to find the velocity of each object from the graphs
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.
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.
Δt =15.0 min
the vehicle’s displacement ()
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.
The vehicle’s displacement is 27.0 km [W].
SC 14.
position-time graph
the average speed (vave), average velocity (), and net displacement ()
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
= (–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.
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.
velocity-time graph
the distance the elk will travel (Δd) in 30 min
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
The elk will travel 18 km.