40 min readNo calculatorNeeds: differentiation, integration
Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — velocity, acceleration and when the particle is at rest
A particle moves so that its displacement after t seconds is s=t3−6t2+9t metres. Find expressions for v and a, and the times at which the particle is instantaneously at rest.
1v=dtds=3t2−12t+9Differentiate the displacement.
2a=dtdv=6t−12Differentiate again.
3At rest means v=0: 3t2−12t+9=0Not s=0 and not a=0.
43(t−1)(t−3)=0, so t=1 and t=3Take the 3 out first, then factorise.
5The particle is at rest after 1 second and after 3 seconds.These are the moments it changes direction.
Example 2 — total distance when the particle turns around
The same particle has s=t3−6t2+9t. Find the total distance it travels in the first 4 seconds, and its displacement at t=4.
1The direction changes at t=1 and t=3, from the previous example.Always find these first — they split the journey into legs.
2s(0)=0, s(1)=1−6+9=4The first leg is 4 m forward.
3s(3)=27−54+27=0The second leg is 4 m backward — back to the start.
4s(4)=64−96+36=4The third leg is 4 m forward again.
5Total distance =4+4+4=12 m, but displacement =s(4)−s(0)=4 m.Add the sizes of the legs for distance; subtract the ends for displacement.
Example 3 — integrating twice, with two conditions
A particle has acceleration a=6t−4 m s−2. When t=0 its velocity is 2 m s−1 and its displacement is 0. Find its displacement when t=2.
1v=∫(6t−4)dt=3t2−4t+cIntegrate the acceleration.
2v=2 at t=0, so c=2 and v=3t2−4t+2Use the velocity condition on the velocity equation.
3s=∫(3t2−4t+2)dt=t3−2t2+2t+kIntegrate again — a second constant appears.
4s=0 at t=0, so k=0Each condition pins down one constant. Do not mix them up.