AP Physics · topic 1
Kinematics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Calculus extension: a reversal hidden inside the interval
A 2.0 kg cart has one-dimensional position x(t) = t³−6t²+9t metres for 0≤t≤4 s, with t in seconds. The supplied position samples are evidence points on this cubic, not straight-line motion between them. No other translational kinetic-energy stores are included. Supplied data (Time / s; Position / m): (0, 0), (1, 4), (2, 2), (3, 0), (4, 4). (a) Differentiate to find velocity, acceleration and net force. Locate reversals and compute total distance and displacement. [3 marks] (b) Calculate net impulse from 0 to 4 s and check net work from 0 to 2 s with kinetic energy. Explain why zero impulse need not mean no force. [3 marks]
Selected position samples on a cubic trajectory
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Position / m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Position / m |
|---|---|
| 0 | 0 |
| 1 | 4 |
| 2 | 2 |
| 3 | 0 |
| 4 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Differentiate to find velocity, acceleration and net force. Locate reversals and compute total distance and displacement.
Reveal the worked solution
- Differentiate x(t): v = dx/dt = 3(t−1)(t−3), and a = dv/dt = 6t−12. Newton's law gives F = ma = 12t−24 N.
- Velocity changes sign at 1 and 3 s. Position values at 0,1,3,4 s are 0,4,0,4 m, so path length is 4+4+4 = 12 m.
- Net displacement is x(4)−x(0) = 4 m. The intermediate t = 2 s sample does not create an additional reversal; smooth cubic behaviour must be used.
- Integrating force gives J = ∫₀⁴(12t−24)dt = [6t²−24t]₀⁴ = 0 N s, matching m[v(4)−v(0)] = 2(9−9).
- At t = 0, v = 9 m s⁻¹ and K = 81 J. At t = 2, v = −3 m s⁻¹ and K = 9 J, so net work is −72 J.
- A direct work integral is ∫F(t)v(t)dt, not ∫F(t)dt. Positive and negative force impulses can cancel even though the force is nonzero during most of the interval.
Answer (a) v = 3t²−12t+9; a = 6t−12; F = 12t−24 N; reversals at 1 and 3 s; distance = 12 m, displacement = 4 m. (b) Net impulse = 0; net work from 0 to 2 s = −72 J.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A moving receiver below a launch platform
A ball leaves a platform horizontally at 6.0 m s⁻¹, 5.0 m above the ground. A receiver starts 4.0 m horizontally from the point directly below the launch and moves away at constant speed 2.0 m s⁻¹. Use g = 10 m s⁻² and neglect air resistance; model both objects as points. Supplied data (Time / s; Receiver horizontal position / m): (0, 4), (0.5, 5), (1, 6), (1.5, 7). (a) Determine whether the ball meets the receiver at ground level and calculate the impact velocity components. [3 marks] (b) The ball is instead launched horizontally at 7.0 m s⁻¹. Find the receiver's required starting position and explain the role of the two independent components. [3 marks]
Receiver position after launch
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Receiver horizontal position / m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Receiver horizontal position / m |
|---|---|
| 0 | 4 |
| 0.5 | 5 |
| 1 | 6 |
| 1.5 | 7 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Determine whether the ball meets the receiver at ground level and calculate the impact velocity components.
Reveal the worked solution
- Vertical motion gives 5.0 = ½(10)t², hence t = 1.0 s.
- The ball reaches x = 6.0t = 6.0 m; the receiver reaches x = 4.0+2.0t = 6.0 m, so they coincide.
- Horizontal velocity stays 6.0 m s⁻¹; vertical velocity is −gt = −10 m s⁻¹. Speed is √136 ≈11.7 m s⁻¹, directed about 59° below horizontal.
- Flight time remains 1.0 s because initial vertical velocity and vertical drop are unchanged.
- The ball reaches x = 7.0 m, while the receiver travels 2.0 m during the fall. Its initial x must therefore be 5.0 m.
- Horizontal speed changes range without changing fall time under this model; a finite receiver or air resistance would require a more detailed tolerance model.
Answer (a) Both reach x = 6.0 m at 1.0 s; impact velocity is (6.0, −10) m s⁻¹. (b) Required starting position = 5.0 m from the launch vertical.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.