2.1.4: Acceleration and Measurement

Illustration 4: Constant Acceleration and Measurement

A 1.0-kg cart on a track experiences several different constant accelerations as shown in the animations (position is given in meters and time is given in seconds). The red dot shows you where position measurements are taken. But how can you tell what acceleration the carts have? There are several ways that we will discuss in this Illustration. Before we begin, play each animation without the velocity calculation (leave it unchecked, since that would be cheating). How would you characterize each cart's motion? How would you describe each cart's acceleration? How can you show that your assertions are correct? Restart.

Hopefully when you looked at Animations 1 and 2 you saw that the carts moved with a constant velocity (Animation 1 has a positive velocity and Animation 2 a negative velocity). The motion of the carts during each animation is uniform, and just by looking you can tell (with practice) that each cart undergoes the same displacement in the same time interval. This is easy enough to check. In Animation 1 the cart is at \(x = 0\text{ m}\) at \(t = 0\text{ s}\), at \(x = 0.5\text{ m}\) at \(t = 0.25\text{ s}\), at \(x = 1.0\text{ m}\) at \(t = 0.5\text{ s}\), at \(x = 1.5\text{ m}\) at \(t = 0.75\text{ s}\), and finally at \(x = 2.0\text{ m}\) at \(t = 1.0\text{ s}\). The cart's motion is uniform at \(v = 2\text{ m/s}\). The cart in Animation 2 has \(v = -2\text{ m/s}\), which can be verified by taking data and showing the velocity calculation for the animation.

What about Animations 3, 4, and 5? Hopefully you saw that the motion was not uniform and that the carts were all accelerating. So how can we prove it and calculate the accelerations? It depends on the situation and the data given. Below are the three most used constant acceleration formulas:

\[v=v_{0}+at\nonumber\]

\[x=x_{0}+v_{0}t+0.5at^{2}\nonumber\]

and

\[v^{2}=v_{0}^{2}+2a(x-x_{0})\nonumber\]

So which of these should we use for Animations 3, 4, and 5? We can rule out the first equation (unless we cheat and turn on the velocity calculation) because it requires instantaneous velocity, which we do not have. We can get position and time measurements from the animation, so that means we can use the second equation. In both Animation 3 and Animation 4 the cart starts at rest, so we have \(x = x_{0} + 0.5at^{2}\), or \(a = 2(x - x_{0})/t^{2}\). The cart has a displacement of \(2\text{ m}\) in \(1\text{ s}\) in Animation 3 and a displacement of \(-2\text{ m}\) in \(1\text{ s}\) in Animation 4. Therefore, the accelerations are \(4\text{ m/s}^{2}\) and \(-4\text{ m/s}^{2}\) respectively.

What about Animation 5? The cart has an initial velocity and slows to a stop (it has a positive velocity and a negative acceleration). How can we calculate this cart's acceleration? We cannot do it with the data given (again without cheating by looking at the velocity data). Why? While it is true that we could estimate the initial velocity by \(\Delta x/\Delta t\), this method will not always yield good results since this is the average velocity during the time interval and not the instantaneous velocity at \(t = 0\text{ s}\). The best way to measure this cart's acceleration is to turn on the velocity calculation and use either \(v = v_{0} + at\) or \(v^{2} = v_{0}^{2} + 2a(x - x_{0})\). This yields an acceleration of about \(-3.7\text{ m/s}^{2}\). Note that if you had used \(\Delta x/\Delta t\) for an estimate of the initial velocity you would have gotten \(3\text{ m/s}\), which differs from the actual \(3.7\text{ m/s}\).

If you leave the velocity calculation on for all of the animations, you can now use \(v = v_{0} + at\) to calculate all of the accelerations.