1. Velocity (s-t) Graphs
This general graph represents the motion of a body travelling at constant velocity. The graph is linear (that is, a straight line).
Recall that linear equations have the general form
y = mx (where m is a constant and x is a variable).
The number m is called the slope of the line (the vertical rise over the horizontal run).
In the above graph, we have the function:
displacement = velocity × time
s = v × t
Velocity is constant and time is a variable.
NOTE: We use the variable "s" for displacement. Be careful not to confuse it with "speed"!
We note that the graph passes through `(0,0)` and has slope v. The slope of the line tells us the velocity. We can also write the velocity using delta notation:
"change in displacement over change in time".
If we have a high velocity, the graph has a steep slope. If we have a low velocity the graph has a shallow slope (assuming the vertical and horizontal scale of each graph is the same).
A marathon runner runs at a constant `12` km/h.
a. Express her displacement travelled as a function of time.
b. Graph the motion for `0 ≤ t ≤ 4\ "h"`
a. `s = 12t`, for displacement s and time t.
b. Graph of `s = 12t`.
We stop the graph at (4, 48).
Please support IntMath!
This is the graph of a journey by sports car:
a. What is the velocity for each stage of the journey?
b. What is the average (mean) velocity for the whole journey?
a. This table outlines the stages of the journey.
|12:00 to 12:30||Travelled `50` km in `30` minutes, so `100` km/h|
|12:30 to 13:00||Stopped|
|13:00 to 14:00||Travelled `50` km in `60` minutes, so `50` km/h|
|14:00 to 14:30||Stopped|
|14:30 to 15:00||Travelled `100` km back towards the starting point in `30` minutes, so `-200` km/h|
b. Even though the whole journey was `200\ "km"` (`100\ "km"` out and `100\ "km"` back) in `3` hours, the displacement for the journey (the distance from the starting point) is `0\ "km"`.
So the average velocity is `0\ "km/h"`.
On the other hand, the average speed was `200/3 = 66.7\ "km/h"`.
`"ave velocity" = "displacement"/"time"`
`"ave speed" = "distance"/"time"`
Get the Daily Math Tweet!
IntMath on Twitter
A particle in a magnetic field moves as follows:
Find the velocity for each part of the motion.
For `t = 0` to `2`: `(Deltas)/(Deltat)=4/2=2\ text(ms)^-1`
For `t = 2` to `7`: `(Deltas)/(Deltat)=0/5=0\ text(ms)^-1`
For `t = 7` to `8`: `(Deltas)/(Deltat)=(-4)/2=-4\ text(ms)^-1`
(The velocity is negative for the last stage, since the particle is going in the opposite direction.)
Please support IntMath!