Acceleration.

Step 1 of 6

A growing velocity

The cart sets off from rest on the start line, and the fan pushes it forward. Every second its velocity grows by 0.2 m/s. How fast the velocity changes is the acceleration:

$a = \frac{\Delta v}{\Delta t}$$a =$$\frac{\Delta v}{\Delta t}$⁠

A change in velocity in metres per second, divided by a time in seconds: the unit is m/s², metres per second every second. Here a = 0.2 m/s².

The position graph is no longer straight: each second the cart covers more ground than in the second before, and the curve gets ever steeper. The orange marks on the track spread further apart.

Step 2 of 6

A cart that is already moving

Now the cart crosses the start line already moving at 0.2 m/s, and the fan adds 0.15 m/s to that every second. The velocity after a time $t$⁠ is the starting velocity plus what has been added:

$v = v_0 + at$⁠

A straight line again: on a velocity–time graph, $v_0$⁠ is the intercept on the vertical axis and the acceleration is the slope. From rest $v_0 = 0$⁠, which leaves $v = at$⁠.

Step 3 of 6

A push against the motion

All three graphs are shown now, one under another on the same time axis: position, velocity and acceleration. The cart crosses the start line at 0.6 m/s, away from the sensor, and the fan is off.

In a moment the fan will come on and push backwards, towards the sensor, with an acceleration of −0.3 m/s², the whole time. Predict what the cart will do.

Step 4 of 6

Negative acceleration, and a cart that speeds up

With a Starting velocity of 0.6 m/s and an Acceleration of −0.3 m/s², the cart slowed down because its velocity and acceleration had opposite signs: the velocity was positive, and the acceleration was making it smaller. After the turn the velocity is negative, and the negative acceleration makes its size grow: the cart speeds up, backwards.

Now the cart is rolling back, towards the sensor, at 0.1 m/s, and the fan is off. In a moment it will come on and push towards the sensor as well, with an acceleration of −0.1 m/s². Predict: will the cart speed up or slow down?

Step 5 of 6

At the turning point

The rule: a cart speeds up when its velocity and acceleration have the same sign, and slows down when their signs are opposite. The acceleration on its own does not say whether the cart is slowing down; only comparing it with the velocity does.

In this run, starting at 0.6 m/s with an acceleration of −0.3 m/s², the cart stops for an instant two seconds in, at the top of the position graph’s arch, and there the velocity graph crosses zero. Look at the acceleration graph at that moment.

Step 6 of 6

Summary

• Acceleration is the rate at which velocity changes, $a = \dfrac{\Delta v}{\Delta t}$$a =$$\dfrac{\Delta v}{\Delta t}$⁠, in m/s².
• With constant acceleration the velocity changes as $v = v_0 + at$⁠: the velocity graph is a straight line and its slope is the acceleration.
• A cart speeds up when its velocity and acceleration have the same sign, and slows down when their signs are opposite. At the turning point the velocity is zero, and the acceleration is not.

Try it: change Starting velocity and Acceleration, and find a run where the cart starts off towards the sensor, stops and comes forward again. What do the signs have to be?

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