Year 11 PhysicsModule 1Module Quiz⏱ ~35 min40 marks
Module 1 Quiz
Kinematics, complete assessment covering both inquiry questions across L01–L08. 15 MC questions (auto-marked) + 5 written questions (self-marked). Complete all questions before submitting.
IQ1
Motion in a Straight Line
IQ1
Graphs & Equations of Motion
IQ2
Motion on a Plane
Written
Extended Problem Solving
Progress
0 / 15 MC answered
Section A, Multiple Choice
15 questions · 1 mark each · 15 marks
Q1, L01 Scalars & Vectors
Which of the following is a vector quantity?
Q2, L01 Vector Addition
Taking east as positive, a person walks 8 m east then 3 m west. The resultant displacement is:
Q3, L02 Distance vs Displacement
A cyclist rides 600 m around a circular track and returns exactly to the starting line. The distance and displacement are:
Q4, L03 Unit Conversion
A car travels at 90 km h⁻¹. Converted to SI units (m s⁻¹), this is:
Q5, L03 Average Speed & Velocity
A swimmer completes 4 lengths of a 25 m pool in 100 s, finishing back at the starting wall. Her average speed and average velocity are:
Q6, L04 Acceleration
A car accelerates uniformly from rest to 20 m s⁻¹ in 5.0 s. Its acceleration is:
Q7, L04 Signed Acceleration
Taking east as positive, a car slows from 20 m s⁻¹ east to 8 m s⁻¹ east in 4.0 s. Its acceleration is:
Q8, L05 Area Under a v–t Graph
On a velocity–time graph, an object starting from rest accelerates uniformly to 8 m s⁻¹ over 4 s. The displacement (triangle area) is:
Q9, L06 Free Fall (SUVAT)
A ball is dropped from rest and falls for 2.0 s. Using g = 9.8 m s⁻², how far does it fall?
Q10, L06 Equation of Motion
A car accelerates uniformly from rest at 3.0 m s⁻² for 4.0 s. How far does it travel?
Q11, L07 Perpendicular Vectors
Two perpendicular velocities of 3 m s⁻¹ and 4 m s⁻¹ act on the same object. The magnitude of the resultant velocity is:
Q12, L07 Displacement on a Plane
A drone flies 6.0 km north then 8.0 km east. The magnitude of its displacement is:
Q13, L07 Relative Velocity
Taking east as positive, car X travels east at 90 km h⁻¹ and car Y travels west at 70 km h⁻¹ directly toward it. The velocity of X relative to Y is:
Q14, L07 River Crossing
A river is 60 m wide. A boat heads straight across at 3.0 m s⁻¹ while the current flows at 4.0 m s⁻¹ downstream. The crossing time is:
Q15, L08 Resultant Displacement
A hiker walks 30 m east then 40 m north. The magnitude of the resultant displacement is:
Section B, Written Questions
5 questions · 5 marks each · 25 marks
Q16, L06: Equations of Motion (SUVAT)5 MARKS
A car starts from rest and accelerates uniformly along a straight road at 2.5 m s⁻² for 8.0 s. (a) Calculate its final velocity. (b) Calculate the distance it travels in this time. (c) Use a second equation of motion to verify your distance in part (b).
Model Answer:(a) v = u + at = 0 + (2.5)(8.0) = 20 m s⁻¹(b) s = ut + ½at² = 0 + ½(2.5)(8.0)² = ½ × 2.5 × 64 = 80 m(c) v² = u² + 2as → 20² = 0 + 2(2.5)s → 400 = 5s → s = 80 m (consistent)Marks: 1, correct equation for v | 1, v = 20 m s⁻¹ | 1, correct equation for s | 1, s = 80 m | 1, valid verification with a second equation
Q17, L06: Vertical Motion Under Gravity5 MARKS
A ball is thrown vertically upward with an initial velocity of 24.5 m s⁻¹. Take up as positive and use g = 9.8 m s⁻² (so a = −9.8 m s⁻²), ignoring air resistance. (a) Calculate the time taken to reach its highest point. (b) Calculate the maximum height reached above the launch point. (c) State the total time of flight until it returns to the launch height, and justify your value.
Model Answer:(a) At the highest point v = 0. v = u + at → 0 = 24.5 + (−9.8)t → t = 24.5 ÷ 9.8 = 2.5 s(b) v² = u² + 2as → 0 = 24.5² + 2(−9.8)s → 0 = 600.25 − 19.6s → s = 600.25 ÷ 19.6 = 30.6 m(c) Total time of flight = 2 × 2.5 = 5.0 s
The upward and downward journeys are symmetric (same distance, same magnitude of acceleration), so the fall back to launch height takes the same 2.5 s as the rise.
Marks: 1, recognise v = 0 at top | 1, t = 2.5 s | 1, correct equation for height | 1, max height = 30.6 m | 1, total flight time 5.0 s with symmetry reason
Q18, L05: Velocity–Time Graph Analysis5 MARKS
A train starts from rest and its motion has three stages: it accelerates uniformly to 12 m s⁻¹ over the first 4.0 s, travels at a constant 12 m s⁻¹ for the next 6.0 s, then decelerates uniformly to rest over a final 2.0 s. (a) Calculate the total displacement over the whole trip using the area under a velocity–time graph. (b) Calculate the acceleration during the first stage. (c) Calculate the acceleration during the final stage.
Model Answer:(a) Stage 1 (triangle) = ½ × 4.0 × 12 = 24 m; Stage 2 (rectangle) = 12 × 6.0 = 72 m; Stage 3 (triangle) = ½ × 2.0 × 12 = 12 mTotal displacement = 24 + 72 + 12 = 108 m(b) a = Δv ÷ Δt = (12 − 0) ÷ 4.0 = 3.0 m s⁻²(c) a = Δv ÷ Δt = (0 − 12) ÷ 2.0 = −6.0 m s⁻² (deceleration)Marks: 1, stage 1 area | 1, stage 2 area | 1, total = 108 m | 1, stage 1 acceleration 3.0 m s⁻² | 1, stage 3 acceleration −6.0 m s⁻²
Q19, L07: Vectors on a Plane, Crosswind5 MARKS
A light plane has an airspeed of 200 km h⁻¹ directed due north. It flies into a wind of 80 km h⁻¹ blowing from the west (that is, toward the east), which acts at right angles to the plane's heading. (a) Calculate the plane's ground speed. (b) Calculate the direction of the ground velocity as an angle east of north. (c) Explain the effect of the crosswind on the plane's actual path over the ground.
Model Answer:(a) The air velocity (north) and wind velocity (east) are perpendicular, so use Pythagoras:v(ground) = √(200² + 80²) = √(40000 + 6400) = √46400 = 215 km h⁻¹ (3 s.f.)(b) θ = arctan(80 ÷ 200) = arctan(0.400) = 21.8° east of north
(c) The eastward wind adds a sideways component to the plane's velocity, so its resultant ground velocity points 21.8° east of north rather than straight north. The plane is carried off its intended heading and its ground speed (215 km h⁻¹) is slightly greater than its airspeed (200 km h⁻¹). To reach a destination due north, the pilot would have to steer slightly into the wind (west of north).
Marks: 1, recognise perpendicular components | 1, ground speed 215 km h⁻¹ | 1, correct trig set-up | 1, angle 21.8° east of north | 1, valid explanation of drift off heading
Q20, L07 & L08: River Crossing, Relative Motion5 MARKS
A river is 80 m wide. A boat is steered so that it heads straight across (perpendicular to the bank) at 2.5 m s⁻¹ relative to the water, while the current flows at 6.0 m s⁻¹ downstream. (a) Calculate the time taken to cross the river. (b) Calculate how far downstream the boat drifts during the crossing. (c) Calculate the boat's resultant speed relative to the bank.
Model Answer:(a) Crossing time depends only on the across-river motion: t = width ÷ v(across) = 80 ÷ 2.5 = 32 s(b) Downstream drift = v(current) × t = 6.0 × 32 = 192 m(c) The across and downstream velocities are perpendicular: v(bank) = √(2.5² + 6.0²) = √(6.25 + 36) = √42.25 = 6.5 m s⁻¹
The current does not change the crossing time, because it acts along the bank while the crossing is governed by the perpendicular (across-river) velocity component.
Marks: 1, use across-river speed for time | 1, t = 32 s | 1, drift = v(current) × t | 1, drift = 192 m | 1, resultant speed 6.5 m s⁻¹ via Pythagoras