Acceleration and displacement as functions of time, and displacement-time graphs

Last time

Before we start: last time, in Constant acceleration, vertical motion and velocity-time graphs, you learned these points. Here they are again on one card. Keep them in mind, because today builds on them.

Calculus meets motion

Displacement, velocity, acceleration: each one is the rate of change of the one before. Differentiate to go down the chain, integrate to go back up. So when a question gives you a formula in t, you have one job: choose the right direction, and use what the question tells you about rest or the start.

The chain of motion

Velocity is the derivative of displacement, and acceleration is the derivative of velocity. Going back, integrate. Velocity is the integral of acceleration, and displacement is the integral of velocity. Each integral brings a constant. Use a given fact, such as starting from rest, to find it.

From acceleration to distance

A particle starts from rest at O with acceleration a equals 6 t minus 18. Integrate: v equals 3 t squared minus 18 t, and the constant is zero because v is zero at the start. Integrate again: s equals t cubed minus 9 t squared, with no constant because s is zero at O. It comes to instantaneous rest when v is zero: 3 t times t minus 6, so t equals 6. Then s of 6 is 216 minus 324, which is minus 108. The distance is 108 metres.