Home > For Students > Velocity and Acceleration

Overview: Now that we’ve delved into the relationship between position and velocity, we certainly can’t forget about acceleration. Whether you’re a car enthusiast or just enjoy taking more derivatives, acceleration is an important concept because it indicates the rate at which velocity is changing! For example, we may know that the velocity is both high and positive, but is the velocity increasing, decreasing, or staying the same? This is where acceleration comes into play. Before we get formal, let’s do a similar exercise to give you an intuitive sense of acceleration.

### Think about which velocity graph has positive acceleration, and then mouse over the image to see if you’re correct!

Misconceptions: As you’ve just seen, acceleration can be pretty intuitive from a graphical perspective. However, when you really think about the graph, things can be a little confusing. For instance, we said that the object on the right was accelerating; all the while, in reality the object is moving to the left, resting, and then moving to the right – how can that be acceleration? Thus, before we continue, we should dispel some common misunderstandings about acceleration in the world of physics. The reason that the object in the second graph is accelerating is because its velocity is consistently increasing – whether positive or negative, it’s increasing. Hence, acceleration can be tricky if you think in layman’s terms! Take a look at the following animation to see a breakdown of the types of acceleration. Show Misconceptions Animation.

Concepts: Now that we’ve dispelled some misconceptions about acceleration, let’s think about it in terms of velocity, and then later we’ll tie it together with position. Formally, acceleration is the rate at which velocity is changing. Hence, if velocity is in terms of meters per second, acceleration is just meters per second per second, or m/s2. Because acceleration is found in the slope of the velocity function, this implies that we can take the derivative of velocity to get acceleration. This means that v’(t) = a(t), where a(t) is the function for acceleration. Finally, in the same way that the critical points of the position function give you possible changes in velocity, the critical points of the velocity function also indicate possible changes in acceleration.

For practice: Find where the acceleration is changing if the velocity of an object is given by v(t) = 3t2 – 4t – 2. Mouse over the image for a quick answer check. If you'd like, you can also see a more detailed explanation. Show/Hide explanation.

• To begin the problem, we note that the velocity of the object is given by v(t) = 3t2 – 4t – 2.
• We want to see where the acceleration is changing sign values, or the critical points of v(t), so we need the acceleration function. To find acceleration from velocity, we take the derivative of v(t). Taking the derivative, we get a(t) = v'(t) = 6t - 4.
• Now, to see where (or if) the acceleration function is changing signs, we need to set a(t) = 0. This gives us 6t - 4 = 0, and so solving we get t = (2/3).
• To see exactly what's happening here, we 'test' values of 't' above and below (2/3). At t = 0, a(t) is negative, and at t = 1, we have that a(t) is positive.
• Because we've found a critical point for v(t), this allows us to conclude that the acceleration changes from negative to positive when t = (2/3).