A.1 · IB Physics
Motion graphs
One car moving in a straight line, three linked pictures. Set the initial velocity and the acceleration, then watch the x–t and v–t graphs draw themselves. Gradient and area carry the meaning.
Predict first
Start from the default settings (u = 4 m s⁻¹, a = 1.0 m s⁻²). Change the acceleration to −2.0 m s⁻². After how long does the car turn round?
The car turns round when v = 0. From v = u + at, 0 = 4 − 2t, so t = u ÷ |a| = 4 ÷ 2 = 2 s.
Check it in the model: Move the Acceleration, a slider from 1.0 to −2.0 m s⁻² and read the Turns round line below the controls.
position x velocity v gradient / shaded area
v = u + at
gradient of x–t = v
gradient of v–t = a
area under v–t = displacement
Try this
- Set a = 0 and u = 8 m s⁻¹. Describe the x–t and v–t graphs, then check that the shaded area under v–t equals the displacement at t = 10 s.
- Set u = 16 m s⁻¹ and a = −3.0 m s⁻². Find the time at which x–t peaks and compare it with the time at which v–t crosses zero.
- Keep u = 4 m s⁻¹ and compare a = +2.0 with a = −2.0 m s⁻². Why do the displacement and the distance travelled differ for the negative value?
Common error: the gradient of x–t is velocity, not speed, and a negative gradient means motion in the negative direction. The area under v–t gives displacement; distance travelled needs the area counted as positive on both sides of the time axis.