The graph showed the relationship with the displacement with the time it took for it to happen. It is hard to get the linear equation. So what we need to do is to trace the graph with the corresponding requirements in the question.
(a) At 10 seconds, the distance is at 11 m.
(b) The displacement is the change in the original position with the current location.
Displacement = final distance – initial distance
Displacement = 11 – 0 = 11m
(c) Velocity = change in distance/ change in time = 11/10 = 1.1 m/s
Next we need to get approximate values on the graph:
Time (x) 2.5 | 5 | 10 |
Distance (y) 5 | 7 | 11 |
Get the value of slope
m = y2-y1 / x2-x1
m = 11-7/10-5 = 0.8
You can get the value of distance at 8 sec (d). You can use any ordered pairs above. In this case we will use (5,7)
Let: Y2 = 7 X2 = 5
Y1 = ? X1 = 8
m = y2-y1 / x2-x1
0.8 = 7-y1 / 5-8
y1 = 9.4
Velocity = 9.4/8 = 1.175 m/s -> (d)
Do the same method for item (e) @ 30 seconds
Let: Y2 = 7 X2 = 5
Y1 = ? X1 = 30
m = y2-y1 / x2-x1
0.8 = 7-y1/5-30
y1 = 27
Velocity = 27/30 = 0.9 m/s -> (e)
See attached image
(a) At 10 seconds, the distance is at 11 m.
(b) The displacement is the change in the original position with the current location.
Displacement = final distance – initial distance
Displacement = 11 – 0 = 11m
(c) Velocity = change in distance/ change in time = 11/10 = 1.1 m/s
Next we need to get approximate values on the graph:
Time (x) 2.5 | 5 | 10 |
Distance (y) 5 | 7 | 11 |
Get the value of slope
m = y2-y1 / x2-x1
m = 11-7/10-5 = 0.8
You can get the value of distance at 8 sec (d). You can use any ordered pairs above. In this case we will use (5,7)
Let: Y2 = 7 X2 = 5
Y1 = ? X1 = 8
m = y2-y1 / x2-x1
0.8 = 7-y1 / 5-8
y1 = 9.4
Velocity = 9.4/8 = 1.175 m/s -> (d)
Do the same method for item (e) @ 30 seconds
Let: Y2 = 7 X2 = 5
Y1 = ? X1 = 30
m = y2-y1 / x2-x1
0.8 = 7-y1/5-30
y1 = 27
Velocity = 27/30 = 0.9 m/s -> (e)
See attached image
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