What this is for. This diagnostic helps you decide which Unit 1 (Kinematics) topics to revisit before you move on. It has one question for each topic, two for the larger Topics 1.2 and 1.3, and three short written questions. These are original Marlbridge practice questions, not past exam questions. The set is not calibrated against real exam results, so it does not give a predicted score. Treat each wrong answer as a pointer to one topic, not as a grade.
How to sit it. Allow about 30 minutes. Close your notes. A calculator is fine for arithmetic and trigonometry (the course exam allows a four-function, scientific or graphing calculator), but do the calculus by hand. Use g = 9.8 m/s² and ignore air resistance; g = 10 m/s² gives answers that round the same way or very close. Positions are in m, velocities in m/s, accelerations in m/s² and t in s, so each numerical coefficient carries the unit that makes its term correct. Every question states its axes.
Question 1 (multiple choice · 1.1)
Take +x east and +y north. Two displacements are A = (2.0 î − 5.0 ĵ) m and B = (−6.0 î + 2.0 ĵ) m. What is the magnitude of A − B?
- (A) 5.0 m
- (B) 10.6 m
- (C) 11.7 m
- (D) 15 m
Answer and explanation
Answer: (B). Subtract component by component: A − B = (2.0 − (−6.0)) î + (−5.0 − 2.0) ĵ = (8.0 î − 7.0 ĵ) m. Its magnitude is √(8.0² + 7.0²) = √113 = 10.6 m.
- (A) is |A + B| = |−4.0 î − 3.0 ĵ| = 5.0 m. It adds the vectors instead of subtracting them.
- (C) adds the magnitudes, |A| + |B| = 5.39 + 6.32 = 11.7 m. Magnitudes add only when two vectors point the same way.
- (D) adds the sizes of the components, 8.0 + 7.0. Perpendicular components combine by Pythagoras.
If you missed this: read “Adding vectors by components” in the Topic 1.1 study guide.
Question 2 (multiple choice · 1.2)
Take +x to the right. A particle moves with x(t) = 4.0t² − 1.0t³ for t ≥ 0. What is its acceleration at the moment (after t = 0) when it is momentarily at rest?
- (A) 0
- (B) +8.0 m/s²
- (C) −8.0 m/s²
- (D) −16 m/s²
Answer and explanation
Answer: (C). v_x = dx/dt = 8.0t − 3.0t², which is zero at t = 8.0/3.0 = 2.67 s. Then a_x = dv_x/dt = 8.0 − 6.0t = 8.0 − 16 = −8.0 m/s².
- (A) assumes that zero velocity means zero acceleration. The velocity is still changing at that instant, from positive to negative.
- (B) is a_x at t = 0, not at the moment the particle stops.
- (D) evaluates only the −6.0t term and drops the constant 8.0.
If you missed this: read “From averages to derivatives” and Worked example 1 in the Topic 1.2 study guide.
Question 3 (multiple choice · 1.2)
Take +x to the right. A particle is at x₀ = +3.0 m at t = 0, and its velocity is v_x(t) = 6.0 − 2.0t. Where is it at t = 5.0 s?
- (A) x = +5.0 m
- (B) x = +8.0 m
- (C) x = +12 m
- (D) x = +13 m
Answer and explanation
Answer: (B). Integrate with the initial condition: x(t) = 3.0 + ∫₀ᵗ (6.0 − 2.0t) dt = 3.0 + 6.0t − 1.0t². At t = 5.0 s, x = 3.0 + 30 − 25 = +8.0 m.
- (A) is the displacement, x(5.0) − x₀ = +5.0 m. The integral gives a change in position; you must add x₀.
- (C) is the position at t = 3.0 s, where v_x = 0 and the particle turns around. It then moves back 4.0 m.
- (D) is the distance travelled, 9.0 m out plus 4.0 m back. Distance is not a position.
If you missed this: read “From derivatives back to position: integration” in the Topic 1.2 study guide.
Question 4 (multiple choice · 1.3)
Take +x to the right. At t = 2.0 s, an object’s position–time graph has a negative slope and curves upward (it is concave up). What is the object doing at that instant?
- (A) Moving in the −x direction and speeding up
- (B) Moving in the −x direction and slowing down
- (C) Moving in the +x direction and slowing down
- (D) Momentarily at rest, with a positive acceleration
Answer and explanation
Answer: (B). The slope of the x–t graph is v_x, so v_x < 0: the object moves in −x. Upward curvature means a_x = d²x/dt² > 0. The velocity and acceleration have opposite signs, so the object slows down.
- (A) reads “curves upward” as “speeding up”. Curvature gives the sign of a_x; you must compare it with the sign of v_x.
- (C) takes the positive acceleration as the direction of motion. The direction of motion is the sign of the slope.
- (D) would need a flat tangent (zero slope). Here the slope is negative.
If you missed this: read “The graph links: slopes down, areas up” in the Topic 1.3 study guide.
Question 5 (multiple choice · 1.3)
A cart starts from rest and speeds up with constant acceleration a. After a distance d its speed is v. In a second run, the cart starts from rest with acceleration a/2 over the same distance d. What is its final speed?
- (A) v/4
- (B) v/2
- (C) v/√2
- (D) v
Answer and explanation
Answer: (C). The equation with no t gives v² = 2ad, so v ∝ √a for a fixed distance. Halving a multiplies v by 1/√2 ≈ 0.71.
- (A) squares the factor instead of taking its square root.
- (B) assumes v ∝ a. That is true for the same time (v = at), but here the distance is fixed and the second run takes longer.
- (D) assumes the final speed depends only on the distance.
If you missed this: read “Functional dependence and factors of change” in the Topic 1.3 study guide.
Question 6 (multiple choice · 1.4)
Take +x east and +y north, with velocities measured relative to still water W. Boat A moves at 6.0 m/s north. Boat B moves at 8.0 m/s east. What is the velocity of B as measured by an observer on A?
- (A) 10 m/s, 37° south of east
- (B) 10 m/s, 37° north of east
- (C) 14 m/s, between east and north
- (D) 2.0 m/s east
Answer and explanation
Answer: (A). v_BA = v_BW + v_WA = v_BW − v_AW = (8.0 î) − (6.0 ĵ) = (8.0 î − 6.0 ĵ) m/s. Magnitude √(8.0² + 6.0²) = 10 m/s; direction tan⁻¹(6.0/8.0) = 37° south of east.
- (B) adds v_AW instead of subtracting it, which flips the north–south component.
- (C) adds the speeds as numbers. The velocities are perpendicular, so combine components.
- (D) subtracts the speeds as numbers, as if the boats moved along one line.
If you missed this: read “Notation: subscripts that name the observer” in the Topic 1.4 study guide.
Question 7 (multiple choice · 1.5)
Take +x horizontal and +y up. A ball is launched over level ground at 20 m/s, 60° above the horizontal. At the highest point of its path, what are its speed and its acceleration?
- (A) Speed 0; acceleration 0
- (B) Speed 0; acceleration 9.8 m/s² downward
- (C) Speed 10 m/s; acceleration 9.8 m/s² downward
- (D) Speed 17 m/s; acceleration 9.8 m/s² downward
Answer and explanation
Answer: (C). Only v_y is zero at the top. v_x never changes, because a_x = 0: v_x = 20 cos 60° = 10 m/s. The acceleration is (0, −9.8) m/s² at every point of the flight.
- (A) and (B) set the whole speed to zero. Only the vertical component vanishes.
- (D) uses 20 sin 60° = 17 m/s, the vertical component at launch, which is exactly the one that has become zero.
If you missed this: read “Projectile motion: a special case” in the Topic 1.5 study guide.
Question 8 (short answer · 1.2)
Take +x forward. A particle passes the origin at t = 0 with v_x0 = +4.0 m/s. Its acceleration is a_x(t) = −1.5t.
(a) Find v_x(t). (b) Find the time when the particle stops. (c) Find its position at that moment. (d) A student uses v_x² = v_x0² + 2a_x Δx with the average acceleration over the stop, −1.73 m/s², and gets 4.6 m. Explain why this is wrong.
Answer and explanation
(a) v_x = 4.0 + ∫₀ᵗ (−1.5t) dt = 4.0 − 0.75t² (m/s).
(b) 4.0 − 0.75t² = 0 gives t² = 5.33, so t = 2.3 s (2.31 s).
(c) x = ∫₀ᵗ (4.0 − 0.75t²) dt = 4.0t − 0.25t³. At t = 2.31 s: x = 9.24 − 3.08 = 6.2 m.
(d) That equation comes from integrating a constant acceleration. Here a_x grows in size with time. The particle keeps most of its speed early on and slows sharply only near the end, so it travels further than a uniform slowdown would. Integrating v_x(t), as in (c), is the only valid method.
If you missed this: work through Worked example 2 in the Topic 1.2 study guide.
Question 9 (short answer · 1.4)
Take +x east and +y north, with all values measured relative to the ground G. Car P drives north at a constant 15 m/s. Car Q drives west at 20 m/s.
(a) Find v_QP, the velocity of Q relative to P, in unit vector notation and as a magnitude and direction. (b) Car Q now brakes. In the ground frame its acceleration is 3.0 m/s² east, while P keeps a constant velocity. State Q’s acceleration as measured from P, and give the reason.
Answer and explanation
(a) v_QP = v_QG + v_GP = v_QG − v_PG = (−20 î) − (15 ĵ) = (−20 î − 15 ĵ) m/s. Magnitude √(20² + 15²) = 25 m/s, at tan⁻¹(15/20) = 37° south of west.
(b) 3.0 m/s² east, the same as in the ground frame. P moves at constant velocity, so P’s frame is inertial. Differentiating v_QP = v_QG − v_PG gives a_QP = a_QG − 0.
If you missed this: read “Converting measurements between frames” in the Topic 1.4 study guide.
Question 10 (short answer · 1.5)
Take +x horizontal and +y up, origin at a nozzle at ground level. A water jet leaves the nozzle at 14 m/s, 40° above the horizontal. It hits a window 3.0 m above the ground while it is coming down.
(a) Find the two components of the launch velocity. (b) Find the time at which the water hits the window. (c) Find the horizontal distance from the nozzle to the window.
Answer and explanation
(a) v_x0 = 14 cos 40° = 10.7 m/s; v_y0 = 14 sin 40° = 9.0 m/s.
(b) y = 9.0t − 4.9t² = 3.0, so 4.9t² − 9.0t + 3.0 = 0. The roots are t = 0.44 s (going up) and t = 1.40 s (coming down). The question says coming down, so t = 1.4 s. Check: the top is at 9.0/9.8 = 0.92 s, and 1.40 s is later.
(c) a_x = 0, so x = 10.7 × 1.40 = 15 m. With g = 10 m/s², you get 1.36 s and 15 m.
If you missed this: read “One motion, two one-dimensional problems” and Worked example 1 in the Topic 1.5 study guide.
Your next step
Mark each question right or wrong. For short answers, count a question as missed if any part went wrong.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 1.1 Scalars and Vectors | 1 | Topic 1.1 study guide |
| 1.2 Displacement, Velocity, and Acceleration | 2, 3, 8 | Topic 1.2 study guide |
| 1.3 Representing Motion | 4, 5 | Topic 1.3 study guide |
| 1.4 Reference Frames and Relative Motion | 6, 9 | Topic 1.4 study guide |
| 1.5 Motion in Two or Three Dimensions | 7, 10 | Topic 1.5 study guide |
How to use your result
- Read the explanation for every question, including the ones you got right. A right answer for a wrong reason is still a gap.
- Missed one question in a topic? Read the named section, then try that topic’s practice set.
- Missed two or more in a topic, or most of a short answer? Work through the whole study guide for that topic, then its practice set and checklist.
- Missed questions across several topics? Start with Topic 1.2. Derivatives, integrals and initial conditions feed every other topic in the unit.
- Got everything right? Go straight to the mixed unit review, where each question combines two or more topics.
Your result here is a guide to what to study next. It does not predict an exam score.