These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This is the algebra-based course, so no calculus is needed. Use g = 9.8 m/s² where gravity appears (exam questions that need a number use 10 m/s², and 9.8 m/s² is also accepted). Ignore air resistance. Round final answers to 2 significant figures unless told otherwise, and keep unrounded values in your calculator until the end. Any scientific calculator is fine.
Question 1 (multiple choice · foundation)
Take +x to the right. A motion diagram records a puck sliding across a rough floor. Its positions at equal 0.50 s intervals are:
| t (s) | 0 | 0.50 | 1.00 | 1.50 | 2.00 |
|---|---|---|---|---|---|
| x (m) | 0 | 1.8 | 3.2 | 4.2 | 4.8 |
Which statement best describes the puck’s motion?
- (A) v_x is positive and a_x is positive, so it speeds up.
- (B) v_x is positive and a_x is negative and constant, so it slows down.
- (C) v_x is negative and a_x is positive, because the gaps get smaller.
- (D) v_x is positive and a_x is negative, with a_x getting larger in size.
Answer and explanation
Answer: (B). x increases every interval, so v_x is positive. The gaps are 1.8, 1.4, 1.0 and 0.6 m, so the puck slows down. The average velocities are 3.6, 2.8, 2.0 and 1.2 m/s. Each falls by 0.8 m/s per 0.50 s, so a_x = −1.6 m/s² in every interval: negative and constant.
- (A) reads the increasing positions as increasing speed. Speed shows in the gaps, not in the positions.
- (C) confuses shrinking gaps with moving backwards. The puck moves left only if x decreases.
- (D) has the right signs, but the gaps shrink by the same 0.4 m each time, so the acceleration is constant, not growing.
Question 2 (multiple choice · core)
Take +x to the right. A drone moving along a straight line has v_x = −2.0 m/s at t = 0. Its acceleration–time graph is made of two flat sections: a_x = +3.0 m/s² from 0 to 2.0 s, then a_x = −1.0 m/s² from 2.0 s to 5.0 s. What is its velocity at t = 5.0 s?
- (A) +3.0 m/s
- (B) +1.0 m/s
- (C) +7.0 m/s
- (D) +4.0 m/s
Answer and explanation
Answer: (B). The area under an a_x–t graph is the change in velocity. Area from 0 to 2.0 s: 3.0 × 2.0 = +6.0 m/s. Area from 2.0 to 5.0 s: −1.0 × 3.0 = −3.0 m/s. Total Δv_x = +3.0 m/s. Final velocity = −2.0 + 3.0 = +1.0 m/s.
- (A) is Δv_x only. It forgets to add the initial velocity of −2.0 m/s.
- (C) counts the area below the axis as positive: −2.0 + 6.0 + 3.0.
- (D) is the velocity at 2.0 s. It stops before the second section of the graph.
Question 3 (multiple choice · core)
Take +y upward. From the same high window, at the same instant, ball P is dropped from rest and ball Q is thrown straight down at 5.0 m/s. Which statement is correct?
- (A) Q has a larger acceleration than P because it was thrown.
- (B) During the first 1.0 s, the velocity of each ball changes by −9.8 m/s.
- (C) Q’s velocity–time graph is steeper than P’s.
- (D) P and Q reach the ground at the same time because they have the same acceleration.
Answer and explanation
Answer: (B). Once released, both balls have a_y = −9.8 m/s². The area under each a_y–t graph for 1.0 s is −9.8 m/s, so each velocity changes by the same amount. Q’s velocity goes from −5.0 to −14.8 m/s; P’s goes from 0 to −9.8 m/s.
- (A) mixes up initial velocity with acceleration. The throw sets v_y0; gravity alone sets a_y.
- (C) is wrong because the slope of a v_y–t graph is the acceleration, which is the same for both. The two lines are parallel, with Q’s starting 5.0 m/s lower.
- (D) ignores the initial velocity. Q is already moving down at the start, so it covers the same distance in less time.
Question 4 (calculation · core)
Take +x in the direction of travel. A cyclist speeds up with constant acceleration from 3.0 m/s to 9.0 m/s over a distance of 48 m. Calculate (a) the acceleration and (b) the time taken.
Worked solution
(a) Time is not known and not needed, so use the equation that leaves out t: v_x² = v_x0² + 2a_x(x − x₀) → 81 = 9.0 + 2a_x(48) → a_x = 72 ÷ 96 = +0.75 m/s².
(b) v_x = v_x0 + a_x t → t = (9.0 − 3.0) ÷ 0.75 = 8.0 s.
Check. For constant acceleration, Δx = ½(v_x0 + v_x)t = ½(3.0 + 9.0)(8.0) = 48 m. ✓
Suggested mark points (3): 1 for choosing an equation without t (or another valid route); 1 for 0.75 m/s² with unit; 1 for 8.0 s.
Common error: starting with v_x = v_x0 + a_x t, which has two unknowns (a_x and t) at this stage. Another valid route is t = 2Δx ÷ (v_x0 + v_x) = 96 ÷ 12 = 8.0 s first, then a_x = 6.0 ÷ 8.0.
Question 5 (graph · core)
Take +y upward, origin at the ground floor. A lift starts from rest. It speeds up uniformly to 3.0 m/s in 2.0 s, moves at a steady 3.0 m/s for 4.0 s, then slows uniformly to rest in 2.0 s.
(a) Sketch the v_y–t graph for the 8.0 s trip, with values on both axes. (b) Sketch the matching a_y–t graph, with values. (c) Find the height of the lift at the end of the trip. (d) Describe the shape of the y–t graph in each stage.
Worked solution and suggested Marlbridge rubric
(a) Three straight segments: from (0 s, 0) up to (2.0 s, 3.0 m/s); flat to (6.0 s, 3.0 m/s); down to (8.0 s, 0).
(b) Slopes of the v_y–t segments: 3.0 ÷ 2.0 = +1.5 m/s² for 0–2.0 s; 0 for 2.0–6.0 s; −3.0 ÷ 2.0 = −1.5 m/s² for 6.0–8.0 s. The a_y–t graph is three flat sections at +1.5, 0 and −1.5 m/s². Check: the areas +3.0 m/s and −3.0 m/s cancel, so the lift ends at rest. ✓
(c) Area under v_y–t: ½ × 2.0 × 3.0 = 3.0 m; 4.0 × 3.0 = 12 m; ½ × 2.0 × 3.0 = 3.0 m. Total 18 m above the ground floor.
(d) 0–2.0 s: curving upward, getting steeper (y = 3.0 m at 2.0 s). 2.0–6.0 s: straight line with slope 3.0 m/s (y = 15 m at 6.0 s). 6.0–8.0 s: still rising but flattening, until it is level at 18 m.
| Point | What earns it |
|---|---|
| 1 | v_y–t sketch with three straight segments and correct values |
| 1 | a_y–t sketch with +1.5, 0 and −1.5 m/s² |
| 1 | Uses areas (or kinematic equations stage by stage) to find each displacement |
| 1 | 18 m with unit |
| 1 | y–t shape: steepening, straight, flattening, with no backward section |
A common error is to draw the y–t graph going down in the last stage because a_y is negative. The lift is still moving up (v_y > 0), so y still increases.
Question 6 (constructed response · stretch)
Take +x in the direction of travel. A driver sees a hazard. For a reaction time t_r the car keeps moving at its initial speed v₀. Then the brakes give a constant acceleration of size a until the car stops.
(a) Derive an expression for the total stopping distance d in terms of v₀, t_r and a. (b) Use t_r = 0.80 s and a = 6.0 m/s² to find d for v₀ = 15 m/s and for v₀ = 30 m/s. (c) A student claims: “Doubling the speed doubles the stopping distance.” Evaluate this claim using your answers. (d) Sketch the v_x–t graph for the 15 m/s case and state what its area represents.
Worked solution and suggested Marlbridge rubric
(a) Reaction stage (constant velocity): d₁ = v₀t_r. Braking stage (a_x = −a, final velocity 0): 0 = v₀² + 2(−a)d₂, so d₂ = v₀² ÷ (2a). d = v₀t_r + v₀² / (2a).
(b) v₀ = 15 m/s: d = 15 × 0.80 + 15² ÷ 12 = 12 + 18.75 = 31 m. v₀ = 30 m/s: d = 30 × 0.80 + 30² ÷ 12 = 24 + 75 = 99 m.
(c) The claim is incorrect. The ratio is 99 ÷ 30.75 ≈ 3.2, not 2. The reaction distance is proportional to v₀, so it doubles (12 m → 24 m). The braking distance is proportional to v₀², so it quadruples (18.75 m → 75 m). The total grows by a factor between 2 and 4.
(d) A flat line at 15 m/s from 0 to 0.80 s, then a straight line down to 0 at 0.80 + 15 ÷ 6.0 = 3.3 s. The area under it (rectangle 12 m plus triangle 18.75 m) is the stopping distance.
| Point | What earns it |
|---|---|
| 1 | Reaction distance v₀t_r |
| 1 | Braking distance v₀² ÷ (2a) from a constant-acceleration equation with correct signs |
| 1 | 31 m and 99 m |
| 1 | Rejects the claim because one term scales with v₀ and the other with v₀² |
| 1 | Sketch with flat then falling straight segments, area identified as d |
Alternative for (a): use the area of the v_x–t graph directly, v₀t_r + ½v₀(v₀ ÷ a). This earns full credit.
Question 7 (constructed response · stretch)
Take +x forward. A car starts from rest at t = 0. Its acceleration is not constant: the a_x–t graph is a straight line falling from 4.0 m/s² at t = 0 to zero at t = 4.0 s. After that, a_x stays zero.
(a) Find the car’s velocity at t = 4.0 s. (b) Describe the shape of the v_x–t graph from 0 to 6.0 s. (c) A student says: “The average acceleration over the first 4.0 s is 2.0 m/s², so the displacement is ½ × 2.0 × 4.0² = 16 m.” Evaluate this claim.
Model answer and suggested Marlbridge rubric
(a) The area under the a_x–t graph is a triangle: ½ × 4.0 s × 4.0 m/s² = 8.0 m/s. Starting from rest, v_x = +8.0 m/s.
(b) From 0 to 4.0 s the graph rises from 0 to 8.0 m/s. It is steepest at the start (slope 4.0 m/s²) and gets less steep, curving over until it is flat at t = 4.0 s, where a_x = 0. From 4.0 to 6.0 s it is a horizontal line at 8.0 m/s.
(c) The claim is incorrect. The equation x = x₀ + v_x0 t + ½a_x t² is valid only for constant acceleration. Using the average value of 2.0 m/s² amounts to replacing the real v_x–t curve with a straight line from 0 to 8.0 m/s, whose area is 16 m. The real curve rises faster at the start, so it lies above that straight line at every time between 0 and 4.0 s. The real area, and so the real displacement, is more than 16 m. It is less than 32 m, because v_x stays below 8.0 m/s until t = 4.0 s.
| Point | What earns it |
|---|---|
| 1 | 8.0 m/s from the area under the a_x–t graph |
| 1 | v_x–t rising, slope decreasing, flat from 4.0 s |
| 1 | States the equation needs constant acceleration |
| 1 | Compares the curve with the straight line (or the areas) to conclude the displacement is more than 16 m |
No numerical value of the true displacement is expected. This course asks you to reason about changing acceleration qualitatively.
How did you do?
- Q1 wrong: re-read “Motion diagrams” and Figure 1 in the study guide.
- Q2 or Q5 wrong: revisit “Graphs: slopes and areas” and Figure 2.
- Q3 wrong: go back to “Free fall near Earth’s surface” and Worked example 2.
- Q4 or Q6 wrong: work through “Where the constant-acceleration equations come from”, Worked example 1 and Worked example 3.
- Q7 incomplete: read “When acceleration is not constant”. Your answer needs the why: a curve above a straight line has more area.
Then tick off the topic checklist.