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 set is for the calculus-based Physics C: Mechanics course. In every formula, x is in m, v_x in m/s, a_x in m/s² and t in s, so each numerical coefficient carries whatever unit makes the term correct. Use g = 9.8 m/s² if needed. Round final answers to 2 significant figures unless told otherwise. A calculator is used only for arithmetic.
Question 1 (multiple choice · foundation)
Take +x to the right. A particle’s position is x(t) = 5.0 + 2.0t − 1.0t³. What is its velocity at t = 2.0 s?
- (A) +0.50 m/s
- (B) −2.0 m/s
- (C) −10 m/s
- (D) −12 m/s
Answer and explanation
Answer: (C). v_x = dx/dt = 2.0 − 3.0t², so v_x(2.0) = 2.0 − 12 = −10 m/s (moving left at 10 m/s).
- (A) divides position by time: x(2.0) = 1.0 m, and 1.0 m ÷ 2.0 s = 0.50 m/s. Position over clock time is not velocity.
- (B) is the average velocity from 0 to 2.0 s: (1.0 − 5.0) m ÷ 2.0 s. It is a secant slope, not the instantaneous value.
- (D) differentiates twice: a_x = −6.0t gives −12 m/s² at 2.0 s. That is the acceleration, with the wrong unit attached.
Question 2 (multiple choice · core)
Take +x to the right. A cart has a_x(t) = 6.0t and an initial velocity v_x0 = −4.0 m/s at t = 0. What is its velocity at t = 2.0 s?
- (A) +8.0 m/s
- (B) +12 m/s
- (C) +16 m/s
- (D) +20 m/s
Answer and explanation
Answer: (A). v_x = v_x0 + ∫₀ᵗ 6.0t dt = −4.0 + 3.0t². At t = 2.0 s: −4.0 + 12 = +8.0 m/s.
- (B) leaves out the initial velocity. The integral gives only the change in velocity (+12 m/s).
- (C) adds the size of v_x0 instead of its signed value: 12 + 4.0. The cart started moving left, so its initial velocity is −4.0 m/s.
- (D) treats the acceleration as constant at its final value, a_x(2.0) = 12 m/s², for the whole 2.0 s: −4.0 + 12 × 2.0. The acceleration grows from 0, so this overestimates the change.
Question 3 (multiple choice · core)
Take +x to the right. A particle has x(t) = t³ − 6.0t². At t = 1.0 s, which statement is correct?
- (A) It is speeding up, because v_x and a_x are both negative.
- (B) It is slowing down, because a_x is negative.
- (C) It is slowing down, because v_x is negative.
- (D) It is neither, because x is negative, so the particle is behind the origin.
Answer and explanation
Answer: (A). v_x = 3.0t² − 12t = −9.0 m/s and a_x = 6.0t − 12 = −6.0 m/s² at t = 1.0 s. Both are negative, so velocity and acceleration point the same way and the speed increases.
- (B) uses the rule “negative acceleration means slowing down”. The sign of a_x alone shows only its direction.
- (C) confuses moving left with slowing down. A negative velocity means motion in −x.
- (D) judges motion from position. x = −5.0 m tells you where the particle is, not how its speed is changing.
Question 4 (calculation · core)
Take +x to the right, with the particle at x₀ = 0 at t = 0. Its velocity is v_x(t) = 3.0t² − 12t + 9.0. Find (a) the displacement and (b) the distance travelled from t = 0 to t = 4.0 s.
Worked solution
- Integrate: x(t) = 0 + ∫₀ᵗ (3.0t² − 12t + 9.0) dt = t³ − 6.0t² + 9.0t.
- (a) Displacement = x(4.0) − x(0) = 64 − 96 + 36 = +4.0 m.
- Turning points: v_x = 3.0(t − 1.0)(t − 3.0) = 0 at t = 1.0 s and 3.0 s. Positions: x(1.0) = 4.0 m, x(3.0) = 0, x(4.0) = 4.0 m.
- (b) Distance = |4.0 − 0| + |0 − 4.0| + |4.0 − 0| = 4.0 + 4.0 + 4.0 = 12 m.
Suggested mark points (3): 1 for x(t) by integration; 1 for the displacement; 1 for splitting at the turning points and adding the sizes.
Common error: reporting ∫₀⁴ v_x dt = 4.0 m as the distance. That is the displacement; the middle leg moves backwards.
Question 5 (constructed response · core)
Take +x along the motion. A particle has constant acceleration a_x. At t = 0 it is at x₀ with velocity v_x0.
(a) Starting from a_x = dv_x/dt, derive an expression for v_x(t). (b) Use your answer to (a) to derive an expression for x(t). (c) Using the chain rule, derive v_x² = v_x0² + 2a_x(x − x₀). (d) State the condition under which all three results hold, and explain which step of your derivation needs it.
Worked solution and suggested Marlbridge rubric
(a) dv_x = a_x dt. Integrate from (0, v_x0) to (t, v_x): v_x − v_x0 = a_x t, so v_x = v_x0 + a_x t.
(b) dx = v_x dt = (v_x0 + a_x t) dt. Integrate from (0, x₀) to (t, x): x = x₀ + v_x0 t + ½a_x t².
(c) a_x = dv_x/dt = (dv_x/dx)(dx/dt) = v_x (dv_x/dx). So a_x dx = v_x dv_x. Integrate: a_x(x − x₀) = ½(v_x² − v_x0²), giving v_x² = v_x0² + 2a_x(x − x₀).
(d) The acceleration must be constant. In each integral, a_x was taken outside the integral sign (∫a_x dt = a_x t; ∫a_x dx = a_x(x − x₀)). That is only valid if a_x does not change with t or x.
| Point | What earns it |
|---|---|
| 1 | (a) Separates and integrates with the initial condition v_x0 |
| 1 | (b) Integrates v_x(t) with the initial condition x₀ to reach the quadratic |
| 1 | (c) Uses the chain rule to write a_x = v_x dv_x/dx |
| 1 | (c) Integrates both sides with correct limits to reach the result |
| 1 | (d) States constant acceleration and identifies taking a_x outside the integral as the step that needs it |
Alternative methods. For (b), indefinite integrals with a constant found from x(0) = x₀ are equally valid. For (c), eliminating t between (a) and (b) algebraically earns both (c) points if every step is shown, since it still relies on constant a_x.
Question 6 (constructed response · stretch)
Take +x along a level track, origin at the start. A fan cart starts from rest at t = 0. The fan speeds up so that the student predicts a_x = ct, where c is a constant. A motion sensor gives:
| t (s) | 0.50 | 1.00 | 1.50 | 2.00 | 2.50 |
|---|---|---|---|---|---|
| x (m) | 0.013 | 0.100 | 0.338 | 0.800 | 1.563 |
(a) Show that the prediction gives x = ct³/6. (b) State what to plot against what to obtain a straight line through the origin if the prediction is correct. (c) Use the data to find c, with its unit. (d) Find the cart’s velocity at t = 2.00 s. (e) Describe a second graph that tests the power of t without assuming it is 3.
Worked solution and suggested Marlbridge rubric
(a) v_x = 0 + ∫₀ᵗ ct dt = ct²/2. Then x = 0 + ∫₀ᵗ (ct²/2) dt = ct³/6.
(b) Plot x (m) on the vertical axis against t³ (s³) on the horizontal axis. The prediction gives a straight line through the origin with slope c/6.
(c) Values of t³: 0.125, 1.00, 3.375, 8.00, 15.625 s³. Each x ÷ t³ is about 0.100 m/s³ (for example 0.800 ÷ 8.00 = 0.100), and a best-fit line through the origin also gives slope 0.100 m/s³. So c = 6 × 0.100 = 0.60 m/s³.
(d) v_x = ct²/2 = 0.60 × (2.00)² ÷ 2 = 1.2 m/s.
(e) If x = kt^n, then ln x = ln k + n ln t. Plot ln x against ln t. The slope gives n; these data give a slope of about 3.0, supporting a_x ∝ t. (Plotting log₁₀ x against log₁₀ t works the same way.)
| Point | What earns it |
|---|---|
| 1 | (a) Two integrations, each using the initial condition (starts from rest at the origin) |
| 1 | (b) x against t³, with the slope identified as c/6 |
| 1 | (c) c = 0.60 m/s³ from the slope, with correct unit |
| 1 | (d) 1.2 m/s from v_x = ct²/2 (carry forward an incorrect c) |
| 1 | (e) Log–log graph with the slope identified as the power of t |
Alternative method for (c). Find average velocities between readings, plot them against midpoint times, and fit v_x = ct²/2. This is valid but less precise because the intervals are long; it earns the point if c is within 0.55–0.65 m/s³.
Question 7 (explanation · stretch)
A student says: “For any motion, the average velocity over an interval equals the mean of the starting and ending velocities, (v_x0 + v_x)/2.”
Take +x along the motion. Test the claim for a particle that starts from rest at the origin with a_x = ct (c is a positive constant), over the interval 0 to T. Then state when the claim is true.
Model answer and suggested Marlbridge rubric
Model answer. From Question 6, v_x = ct²/2 and x = ct³/6.
- True average velocity: v_avg = Δx/Δt = (cT³/6) ÷ T = cT²/6.
- Mean of end velocities: (0 + cT²/2)/2 = cT²/4.
These differ, so the claim is false in general: the true average is only 2/3 of the student’s value. The particle moves slowly for most of the interval and fast only near the end, so the time-weighted average is lower than the mean of the end values.
The claim is true when acceleration is constant. Then v_x is a linear function of t, the v_x–t graph is a straight line, and the area under it (the displacement) equals Δt × (v_x0 + v_x)/2, the area of a trapezium.
| Point | What earns it |
|---|---|
| 1 | Finds the true average velocity cT²/6 from Δx/Δt |
| 1 | Finds the mean of end velocities cT²/4 and concludes the claim fails |
| 1 | Explains the difference physically or graphically (more time spent at low speed; curve below the straight chord) |
| 1 | States the claim holds for constant acceleration, with the trapezium or linear-v_x reason |
A numerical test with a chosen c and T (for example c = 0.60 m/s³ and T = 2.0 s gives 0.40 m/s against 0.60 m/s) earns the first two points.
How did you do?
- Q1 or Q3 wrong: re-read “From averages to derivatives” and the sign chart in Worked example 1 of the study guide.
- Q2 or Q4 wrong: revisit “From derivatives back to position” and Worked example 2. Add initial conditions; split at v_x = 0 for distance.
- Q5 incomplete: go through “Deriving the constant-acceleration equations” step by step.
- Q6 or Q7 incomplete: your reasoning must say why a graph is straight or why an average differs. Compare with Figure 2 and the failure of constant-a equations in Worked example 2.
Then tick off the topic checklist.