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 v_{x}.
(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. v_{x} 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

v_{x} = dx/dt, so
x = ∫v_{x}dt

In this example, we are assuming v_{x} is constant, so the integral becomes

x = v_{x}t + 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 x_{0}
as the initial value of x, so

x_{0} = v_{x}.0 + constant

so the constant is x_{0} and we have the complete equation for x(t):

x = v_{x}t + x_{0}

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

a_{x} = dv_{x}/dt, so
v_{x} = ∫a_{x}dt

In this example, we are assuming a_{x} is constant, so the integral becomes

v_{x} = a_{x}t + constant'

where the dash has been added because this constant is different from that used above. Here we need one value of (t,v_{x}) find the constant and again we shall use the initial condition: here, let's take define v_{x}_{0}
as the initial value of v_{x}, so

v_{x}_{0} = a_{x}.0 + constant'

so the complete equation for v_{x}(t) is

v_{x} = a_{x}t + v_{x}_{0} (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 = ∫v_{x}dt and substituting from the equation above
= ∫a_{x}t + v_{x}_{0}dt
= ½a_{x}t^{2} + v_{x}_{0}t + constant"

Again, we define x_{0} as the initial value of x, so

x = ½a_{x}t^{2} + v_{x}_{0}t + x_{0} (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

Multiplying both sides by 2a_{x} and simplifying gives:

2a_{x}(x − x_{0}) = v_{x}^{2} − v_{x}_{0}^{2} (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.