Linking the graphs: the ball thrown upwards

Throw a ball straight up and catch it. Simple? It's also the examiners' favourite trap: "What is the acceleration at the very top?" Half the room says zero. Let's make sure you're in the other half.

The three graphs are linked in a chain:

$s\text{–}t \xrightarrow{\ \text{gradient}\ } v\text{–}t \xrightarrow{\ \text{gradient}\ } a\text{–}t$

Going back up the chain, the area under the velocity–time graph gives the displacement.

• Displacement–time: a curve. It's steep at first (fast), flattens to a peak (the top, where the gradient and so $v$ are zero), then gets steep again on the way down.
• Velocity–time: a straight line from $+u$ to $-u$. It crosses zero at the top. Its gradient is $-9.81\ \text{m s}^{-2}$ all the way.
• Acceleration–time: a horizontal line at $-9.81\ \text{m s}^{-2}$. Gravity doesn't take a break at the top.

The velocity–time graph is also symmetrical: the ball takes as long to come down as to go up, and it's caught at the speed it was thrown.