Unit 4 / Practice Drills
Unit 4 Practice Drills
Kinematics
The motion-description unit. Projectile and rectilinear problems are the safest scorers; variable acceleration is the trap.
Unit 4 Practice Drills - Kinematics
5 drills covering rectilinear motion, projectile motion, velocity-time graphs, and normal-tangential acceleration.
Work each setup first, then open the steps one by one. The figures are drawn to make the motion and geometry visible before you start substituting into formulas.
Drill 1 - Rectilinear Motion: x = f(t)
Problem statement
The displacement of a particle moving along a straight line is given by x = 2t³ - 9t² + 12t - 4 metres, where t is in seconds.
Find: (a) velocity and acceleration as functions of time, (b) when the particle is momentarily at rest, (c) position, velocity, and acceleration at t = 3 s, (d) total distance travelled in the first 3 seconds.
Figure 1A
Number line with reversals and blank value table
Q1 - Differentiate x(t) to get v(t) and a(t).
x(t) = 2t3 - 9t2 + 12t - 4
v(t) = dx/dt = 6t2 - 18t + 12 = 6(t - 1)(t - 2) m/s
a(t) = dv/dt = 12t - 18 m/s²
Q2 - Set v(t) = 0 and find when the particle is momentarily at rest.
6(t - 1)(t - 2) = 0
t = 1 s or t = 2 s
These are the two instants when the particle stops and reverses direction.
In the first 3 seconds there are two such moments.
Q3 - Substitute t = 3 s into x, v, and a.
x(3) = 2(3)3 - 9(3)2 + 12(3) - 4 = 54 - 81 + 36 - 4 = 5 m
v(3) = 6(3)2 - 18(3) + 12 = 54 - 54 + 12 = 12 m/s
a(3) = 12(3) - 18 = 18 m/s²
Q4 - Find x at t = 0 and at each reversal time, then mark the motion direction.
x(0) = -4 m
x(1) = 2 - 9 + 12 - 4 = 1 m
x(2) = 16 - 36 + 24 - 4 = 0 m
From t = 0 to 1 s, v > 0 so the particle moves forward from -4 m to 1 m.
From t = 1 to 2 s, v < 0 so it moves backward from 1 m to 0 m.
From t = 2 to 3 s, v > 0 again so it moves forward from 0 m to 5 m.
Q5 - Add the absolute lengths of each segment to get total distance.
Distance from -4 m to 1 m = 5 m
Distance from 1 m to 0 m = 1 m
Distance from 0 m to 5 m = 5 m
Total distance = 5 + 1 + 5 = 11 m
The displacement over the same interval is only 5 - (-4) = 9 m, so distance and displacement are not the same here.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Worked Diagram Hidden
Reveal worked diagram
Open this only after you have attempted the setup yourself.
Figure 1B
Worked number line with reversal points and filled values
Where students lose marks
Students often stop after finding the displacement over 0 to 3 s and call it total distance. This drill is about sign changes: once the velocity changes sign, distance must be added segment by segment using absolute values.
The rest of Unit 4 stays premium
The rectilinear motion drill stays free as the sampler. The projectile, wall-clearance, v-t graph, and curvilinear acceleration drills unlock with the premium pass.