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.
Drill 2 - Projectile: 45 m/s at 20°
Problem statement
A ball is projected from ground level with an initial speed of 45 m/s at 20° above the horizontal.
Find: (a) horizontal and vertical components of initial velocity, (b) maximum height reached, (c) total time of flight, (d) horizontal range, (e) velocity (magnitude and direction) at t = 2 s.
Figure 1A
Projectile with components, apex, range, and velocity at t = 2 s
Q1 - Calculate the initial velocity components.
ux = 45 cos 20° = 42.29 m/s
uy = 45 sin 20° = 15.39 m/s
Q2 - Use vy² = uy² - 2gH to find maximum height.
At the apex, vy = 0
0 = (15.39)2 - 2(9.81)H
H = (15.39)2 / 19.62 = 12.07 m
Q3 - Set y = uy t - 1/2 g t² = 0 to find total time of flight.
0 = 15.39t - 4.905t2
t(15.39 - 4.905t) = 0
t = 0 or T = 15.39 / 4.905 = 3.14 s
The root t = 0 is the launch instant, so the non-zero root is the total time of flight.
Q4 - Find horizontal range and explain why ux stays constant.
R = uxT = 42.29 x 3.14 = 132.73 m
The horizontal component stays constant because, neglecting air resistance, there is no horizontal acceleration.
Q5 - Find the velocity at t = 2 s and decide whether the ball is rising or falling.
vx = ux = 42.29 m/s
vy = uy - gt = 15.39 - 9.81(2) = -4.23 m/s
v = √(42.29² + 4.23²) = 42.50 m/s
Angle = tan-1(4.23 / 42.29) = 5.7° below the horizontal
Because the vertical component is negative, the ball is already falling at t = 2 s.
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 projectile with solved height, range, and t = 2 s velocity
Where students lose marks
The two repeated errors are failing to resolve the launch speed into components first and mishandling the y = 0 equation by dividing away the trivial root t = 0. Keep horizontal and vertical motion separate.
Drill 3 - Projectile: 50 m/s at 25° (Wall Clearance)
Problem statement
A stone is projected from ground level at 50 m/s at 25° to the horizontal. A wall 18 m high stands 85 m away horizontally.
Does the stone clear the wall? Find the maximum height and horizontal range on level ground.
Figure 1A
Wall-clearance projectile on level ground
Q1 - Find ux, uy, and the time taken to reach the wall.
ux = 50 cos 25° = 45.32 m/s
uy = 50 sin 25° = 21.13 m/s
twall = 85 / 45.32 = 1.88 s
Q2 - Find the height at x = 85 m.
y = uytwall - 1/2 gtwall2
y = 21.13(1.88) - 4.905(1.88)2
y = 22.37 m
Q3 - Compare the height at the wall with 18 m.
Clearance = 22.37 - 18 = 4.37 m
The stone clears the wall by 4.37 m.
Q4 - Find total time of flight and horizontal range.
T = 2uy / g = 2(21.13) / 9.81 = 4.31 s
R = uxT = 45.32 x 4.31 = 195.2 m
Q5 - Find maximum height and where the apex occurs.
H = uy2 / 2g = (21.13)2 / 19.62 = 22.76 m
tH = uy / g = 2.154 s
xH = uxtH = 45.32 x 2.154 = 97.6 m
The apex is after the wall because 97.6 m is greater than 85 m.
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 wall-clearance result at x = 85 m
Where students lose marks
Use the time to the wall from horizontal motion first, then substitute that same time into the vertical equation. Students who substitute the full flight time into the wall-height calculation answer a different question.
Drill 4 - Uniform Acceleration: v-t Graph
Problem statement
A train starts from rest and accelerates uniformly at 1.8 m/s² for 15 seconds. It then moves at constant speed for 30 seconds. It finally decelerates uniformly and comes to rest in 10 seconds.
Find: (a) maximum speed, (b) deceleration during braking, (c) total distance covered, (d) average speed for the entire journey.
Figure 1A
Velocity-time graph with three phases and shaded area
Q1 - Find the maximum speed at the end of Phase 1.
vmax = 0 + 1.8(15) = 27 m/s
This is the constant speed during Phase 2.
Q2 - Find the braking deceleration in Phase 3.
0 = 27 - a3(10)
a3 = 27 / 10 = 2.7 m/s²
Q3 - Find the distance in each phase from the v-t graph area.
s1 = 1/2 x 15 x 27 = 202.5 m
s2 = 30 x 27 = 810 m
s3 = 1/2 x 10 x 27 = 135 m
Q4 - Add the three areas to get total distance.
Total distance = 202.5 + 810 + 135 = 1147.5 m
Q5 - Find average speed and explain why it is not the simple average of phase speeds.
Total time = 15 + 30 + 10 = 55 s
Average speed = 1147.5 / 55 = 20.86 m/s
Average speed is total distance divided by total time.
It is not the simple average of the phase speeds because the train spends different lengths of time in each phase, and the accelerating and braking phases do not stay at one fixed speed.
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 v-t graph with all distances labeled
Where students lose marks
The distance comes from the area under the v-t graph, not from one kinematic equation across the full trip. Also, average speed is total distance divided by total time, not the simple mean of a few listed speeds.
Drill 5 - Curvilinear: Normal and Tangential Components
Problem statement
A car travels along a curved road of radius of curvature 150 m at a speed of 72 km/h. At this instant the car is also accelerating along the path at 2 m/s².
Find: (a) tangential acceleration, (b) normal acceleration, (c) total acceleration (magnitude and direction).
Figure 1A
Normal and tangential acceleration on a curved path
Q1 - Convert the speed into m/s.
v = 72 / 3.6 = 20 m/s
Q2 - State the tangential acceleration and its meaning.
at = 2 m/s²
Tangential acceleration acts along the path. Because it is in the direction of motion here, the car is speeding up.
Q3 - Calculate the normal acceleration and explain what causes it.
an = v2 / ρ = 20² / 150 = 400 / 150 = 2.67 m/s²
Normal acceleration is caused by the change in direction of velocity and always points toward the centre of curvature.
Q4 - Find the magnitude of total acceleration.
a = √(at2 + an2)
a = √(2² + 2.67²) = √(4 + 7.11) = √11.11 = 3.33 m/s²
Q5 - Find the direction of the total acceleration from the normal axis.
tan β = at / an = 2 / 2.67 = 0.75
β = tan-1(0.75) = 36.9°
Because the normal component is larger than the tangential component, the total acceleration points more toward the centre than along the path.
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 acceleration triangle for normal and tangential components
Where students lose marks
Convert 72 km/h before doing anything else. After that, keep normal and tangential acceleration separate until the final triangle; mixing them too early is what produces the wrong direction angle.