9.7 Distance­ Time Graphs
Apr 15­9:10 PM
y = mx + b
y is the dependent variable
x is the independent variable
m is the slope of line
b is the y intercept of the line
d = v t + 0
d (distance travelled) is the dependent variable
t (time) is the independent variable
v (speed) is the slope of the line
0 (initial distance) is the y intercept
Apr 15­11:58 AM
1
slope = rise
run
v = d
t
Apr 15­11:58 AM
The slope of the line of a distance­time graph for constant speed is related to the defining equation for speed; slope = rise/run is the same as vav = d/ t
Apr 15­9:19 PM
2
Questions
Page 365, #1­5
Apr 15­9:11 PM
1.
2.
Explain why a graph is sometimes more useful than an equation.
What does the slope of a distance­time graph represent?
Apr 24­9:17 AM
3
3.
What interpretation can be made about a moving car if the line on
a distance­time graph for the car has the following characteristics?
a) A high, or steep slope?
Apr 24­9:21 AM
b)
A low, or less steep slope?
Apr 24­9:23 AM
4
c)
A zero slope?
(what does a zero slope look like?)
Apr 24­9:24 AM
d)
A short line on the graph?
Apr 24­9:28 AM
5
e)
A long line on the graph?
Apr 24­9:30 AM
What would this graph tell you about the movement of an object?
Apr 24­9:31 AM
6
4.
Sketch a distance­time graph for a car cruising at 80 km/h
*First, create a table of values.
Time (hours)
Distance (km)
0
80
160
240
320
400
480
560
640
0
1
2
3
4
5
6
7
8
Apr 24­9:32 AM
Distance (km)
Distance­time graph of a car travelling 80 km/h
640
560
480
400
320
240
160
80
Time (hours)
1
2
3
4
5
6
7
8
Apr 24­9:38 AM
7
5. A car leaves Borden­Carleton PEI, on its way across the Confederation Bridge into New brunswick. The distances and times from the toll booth in PEI are in the table. a) Plot a distance time graph using the information
Nov 18­9:46 PM
Distance (km)
Car Crossing Confederation Bridge
16
14
12
10
8
6
4 2
Time (mins)
2
4
6
8
10
12
14
16
Apr 24­9:38 AM
8
B) According to the graph the distance travelled at the end of 5 mins is 6.0 km
C) According to the graph the time required to cross over the 12.9 km bridge is 10.8 mins
D) The speed was constant because the car travelled equal distances in equal time intervals
Nov 18­10:33 PM
E) rise = d = (1.44­0)km = 14.4km
run = t = (12.0 ­0) min = 12.0 min
slope = rise
run
v = d 14.4 km t 12.0 min
= 1.2 km
min
The slope of the graph is 1.20 km min
This slope represents the average speed for the car crossing Confederation Bridge
Nov 18­10:37 PM
9
F) Vav = 1.20 km X 60 min = 72.0 km min
1h h
Nov 18­10:37 PM
10

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