Kinematics and displacement-time graphs – background material for Physclips
Displacement-time graphs are used in kinematics, the quantitative study of motion. In kinematics, the motion is quanified, but not explained: forces, energy and momenta come later. This page presents some supporting material for the multimedia tutorial about constant acceleration from Physclips.
Let's consider motion in the x direction with constant velocity vx.
(Yes, in one dimension we don't really need the subscript, but we'll need it later when we look at projectiles, so let's get used to it here.) The animation shows a man walking in the x direction with constant velocity, and we plot his velocity and position as functions of time. vx is positive, which means that he is walking in the positive x direction, so his x coordinate increases in time. (If these plots aren't clear, see the page on Graphs, errors, significant figures, dimensions and units.)
Now let's analyse it. The definition of velocity is the rate of change in displacement with respect to time, v = dr/dt . So here, in the x direction, we can write
vx = dx/dt, so
x = ∫vxdt
In this example, we are assuming vx is constant, so the integral becomes
x = vxt + constant
How to determine the constant of integration?
If we know one value of (t,x) we can substitute into this equation to find the constant. Quite often, the value that we know is the initial condition, i.e. the value of x when t = 0. Here, let's take define x0
as the initial value of x, so
x0 = vx.0 + constant
so the constant is x0 and we have the complete equation for x(t):
x = vxt + x0
which is the equation graphed in the animation above. Because the velocity is constant, we can use a triangle to determine the slope of this straight line graph and thus obtain the velocity from the graph of x(t), as shown.
Constant acceleration
Now let's consider the case with constant acceleration, here positive. The man is still walking to the right, but his velocity is increasing with time, as the top graph shows.
Again, let's analyse it. The definition of acceleration is the rate of change in velocity with respect to time, a = dv/dt . So here, in the x direction, we can write
ax = dvx/dt, so
vx = ∫axdt
In this example, we are assuming ax is constant, so the integral becomes
vx = axt + constant'
where the dash has been added because this constant is different from that used above. Here we need one value of (t,vx) find the constant and again we shall use the initial condition: here, let's take define vx0
as the initial value of vx, so
vx0 = ax.0 + constant'
so the complete equation for vx(t) is
vx = axt + vx0 (1)
which is the equation graphed in the animation above. Because the velocity is constant, we can use a triangle to determine the slope of this straight line graph and thus obtain the velocity from the graph of x(t), as shown.
To obtain an expression for x(t) in this case, we may once again use the definition of velocity, whence
x = ∫vxdt and substituting from the equation above
= ∫axt + vx0dt
= ½axt2 + vx0t + constant"
Again, we define x0 as the initial value of x, so
x = ½axt2 + vx0t + x0 (2)
which is the parabolic graph plotted above.
What if we need to relate x, a and v without using t? In that case, we can rearrange equation (1) above to give
t = (vx − vx0)/ax
and substitute into the equation (2):
x = ½(vx − vx0)2/ax + vx0(vx − vx0)/ax + x0
Multiplying both sides by 2ax and simplifying gives:
2ax(x − x0) = vx2 − vx02 (3)
Equations 1, 2 and 3 are rather useful and so it's worth remembering them, rather than having to derive them every time you need them.
Now, if you'd now like to look at motion in two dimensions, you can proceed to projectiles and circular motion.
