Variable Acceleration (Calculus) & 2D Projectile Motion
In A-Level Mechanics, the first question you must ask before writing any equation is: 'Is the acceleration constant or a function of time t?' When acceleration varies with time, whether in 1D along a straight line or in 2D using i-j vectors, the SUVAT equations are completely invalid and you must use differentiation and integration. Conversely, a projectile moving freely under gravity has constant acceleration (0 horizontally and -g vertically), allowing you to combine horizontal and vertical SUVAT equations to find maximum height, time of flight, and the parabolic Cartesian trajectory equation.
- Variable Acceleration in 1D: The s-v-a Calculus Ladder
- When displacement s, velocity v, or acceleration a is a function of time t, move down the ladder by differentiating with respect to t: v = ds/dt and a = dv/dt = d²s/dt². Move up the ladder by integrating with respect to t: v = ∫ a dt + c₁ and s = ∫ v dt + c₂. Always use initial conditions (such as 'starts from rest at the origin', meaning v = 0 and s = 0 when t = 0) to find the constants of integration c₁ and c₂, or use definite integration where displacement change from t = t₁ to t = t₂ is Δs = ∫_(t₁)^(t₂) v dt.
- Instantaneous Rest, Maximum Speed, and Total Distance
- 1. Instantaneous rest / changing direction of motion: occurs when velocity v = 0 (so ds/dt = 0). 2. Maximum or minimum velocity: occurs when acceleration a = dv/dt = 0. 3. Total distance travelled vs displacement on [0, T]: if the particle changes direction (v = 0) at some time t = t₁ inside (0, T), the single integral ∫₀^T v dt gives the net displacement (because backward motion subtracts from forward motion). To find total distance travelled, you must split the integral at every root of v(t) = 0 and sum the magnitudes: Distance = |∫₀^(t₁) v dt| + |∫_(t₁)^T v dt|.
- Variable Acceleration in 2D Vectors
- For a particle moving in a plane with position vector r(t) = x(t)i + y(t)j, differentiate or integrate the i and j components independently with respect to t: velocity is v = dr/dt = (dx/dt)i + (dy/dt)j (often denoted ṙ), and acceleration is a = dv/dt = (d²x/dt²)i + (d²y/dt²)j (denoted r̈). Remember to add a vector constant of integration c = c₁i + c₂j when integrating. Speed is the scalar magnitude of the velocity vector: Speed = |v| = √[(dx/dt)² + (dy/dt)²].
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Variable Acceleration in 1D: The s-v-a Calculus Ladder
- When displacement s, velocity v, or acceleration a is a function of time t, move down the ladder by differentiating with respect to t: v = ds/dt and a = dv/dt = d²s/dt². Move up the ladder by integrating with respect to t: v = ∫ a dt + c₁ and s = ∫ v dt + c₂. Always use initial conditions (such as 'starts from rest at the origin', meaning v = 0 and s = 0 when t = 0) to find the constants of integration c₁ and c₂, or use definite integration where displacement change from t = t₁ to t = t₂ is Δs = ∫_(t₁)^(t₂) v dt.
- Instantaneous Rest, Maximum Speed, and Total Distance
- 1. Instantaneous rest / changing direction of motion: occurs when velocity v = 0 (so ds/dt = 0). 2. Maximum or minimum velocity: occurs when acceleration a = dv/dt = 0. 3. Total distance travelled vs displacement on [0, T]: if the particle changes direction (v = 0) at some time t = t₁ inside (0, T), the single integral ∫₀^T v dt gives the net displacement (because backward motion subtracts from forward motion). To find total distance travelled, you must split the integral at every root of v(t) = 0 and sum the magnitudes: Distance = |∫₀^(t₁) v dt| + |∫_(t₁)^T v dt|.
- Variable Acceleration in 2D Vectors
- For a particle moving in a plane with position vector r(t) = x(t)i + y(t)j, differentiate or integrate the i and j components independently with respect to t: velocity is v = dr/dt = (dx/dt)i + (dy/dt)j (often denoted ṙ), and acceleration is a = dv/dt = (d²x/dt²)i + (d²y/dt²)j (denoted r̈). Remember to add a vector constant of integration c = c₁i + c₂j when integrating. Speed is the scalar magnitude of the velocity vector: Speed = |v| = √[(dx/dt)² + (dy/dt)²].
- 2D Projectile Motion: Horizontal and Vertical Resolution
- For a projectile launched at speed U at an angle α above the horizontal (modelling the object as a particle so air resistance is negligible): 1. Horizontally (a_x = 0, constant velocity): u_x = U cos α, so v_x = U cos α at all times and horizontal displacement is x = (U cos α)t. 2. Vertically (taking upwards as positive, constant acceleration a_y = -g): u_y = U sin α, v_y = U sin α - gt, and vertical displacement from launch height is y = (U sin α)t - (1/2)gt².
- Maximum Height, Time of Flight, and Horizontal Range
- 1. At Maximum Height: the vertical component of velocity is momentarily zero (v_y = 0, though total speed is not zero because v_x = U cos α!). Setting v_y = U sin α - gt = 0 gives time to peak t_peak = (U sin α) / g, and using v_y² = u_y² + 2a_y y gives H_max = (U² sin²α) / (2g). 2. Time of Flight T (landing on horizontal ground at the same level, y = 0): set y = (U sin α)t - (1/2)gt² = 0 → T = (2U sin α) / g. 3. Horizontal Range R on level ground: R = u_x × T = (U² sin 2α) / g (which is maximum when 2α = 90° → α = 45°).
- Deriving the Cartesian Equation of the Trajectory
- Never memorise the trajectory formula, exam questions always ask you to prove it from first principles. First write the parametric equations x = (U cos α)t and y = (U sin α)t - (1/2)gt². Rearrange the horizontal equation for t: t = x / (U cos α). Substitute this expression for t into the vertical equation: y = (U sin α)[x / (U cos α)] - (1/2)g[x / (U cos α)]² = x tan α - [gx² / (2U² cos²α)]. Since 1 / cos²α ≡ sec²α ≡ 1 + tan²α, this can also be written as a quadratic in tan α: y = x tan α - [gx² / (2U²)](1 + tan²α).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
In a variable acceleration question, what mathematical condition corresponds to (a) the particle changing direction, and (b) the particle reaching maximum velocity?
Why can ∫₀^5 v(t) dt give a smaller answer than the total distance travelled in the first 5 seconds?
What is the speed of a projectile at the highest point of its trajectory?
How do you find the angle to the horizontal at which a projectile is moving at time t?
What trigonometric identity is used to convert 1 / cos²α into an expression involving tan α in the projectile trajectory equation?
If a projectile is launched from the top of a cliff of height 40 m above the sea, what value of y (vertical displacement) do you substitute to find when it hits the sea?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A particle P moves in a straight line such that its velocity v m s⁻¹ at time t seconds (t ≥ 0) is given by v = 3t² - 18t + 24. When t = 0, P is at the origin O. Find: (a) the values of t when P is instantaneously at rest, (b) the minimum velocity of P, and (c) the total distance travelled by P in the interval 0 ≤ t ≤ 5. (7 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Set v = 0 for instantaneous rest: 3t² - 18t + 24 = 0 → 3(t² - 6t + 8) = 3(t - 2)(t - 4) = 0 → t = 2 s and t = 4 s [2 marks]. 2. (b) Differentiate v to find acceleration a = dv/dt = 6t - 18; set a = 0 → t = 3 s, and substitute t = 3 into v: v_min = 3(3)² - 18(3) + 24 = -3 m s⁻¹ [2 marks]. 3. (c) Integrate v with respect to t to find displacement s(t) (with s(0) = 0 so c = 0): s(t) = t³ - 9t² + 24t [1 mark]. 4. Since P reverses direction at t = 2 and t = 4, evaluate s at t = 0, 2, 4, and 5: s(0) = 0, s(2) = 8 - 36 + 48 = 20 m, s(4) = 64 - 144 + 96 = 16 m, and s(5) = 125 - 225 + 120 = 20 m [1 mark]. 5. Sum the distances of the three segments: Distance = |20 - 0| + |16 - 20| + |20 - 16| = 20 + 4 + 4 = 28 m [1 mark].
- Q2
A particle of mass 2 kg moves in the x-y plane under the action of a single variable force F = (6t i + (12 - 6t²) j) N, where t is the time in seconds (t ≥ 0). Given that when t = 0, the particle is at the origin O and has velocity (2i - 4j) m s⁻¹, find the speed of the particle when t = 2 and its distance from O when t = 2. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Apply Newton's Second Law a = F / m with m = 2 kg: a = (3t)i + (6 - 3t²)j m s⁻² [1 mark]. 2. Integrate acceleration with respect to t to find velocity v: v = ∫ a dt = (3/2 t² + c₁)i + (6t - t³ + c₂)j [1 mark]. 3. Use v(0) = 2i - 4j to find c₁ = 2 and c₂ = -4, so v = (3/2 t² + 2)i + (6t - t³ - 4)j [1 mark]. 4. Evaluate v at t = 2: v(2) = (6 + 2)i + (12 - 8 - 4)j = 8i + 0j m s⁻¹, so Speed = |v(2)| = 8 m s⁻¹ [1 mark]. 5. Integrate v with respect to t (with r(0) = 0i + 0j) to find position vector r: r = (1/2 t³ + 2t)i + (3t² - 1/4 t⁴ - 4t)j [1 mark]. 6. Evaluate r at t = 2: r(2) = (4 + 4)i + (12 - 4 - 8)j = 8i + 0j m, so distance from O is |r(2)| = 8 m [1 mark].
- Q3
A ball is projected from a point A at a height of 15 m above horizontal ground with initial speed 25 m s⁻¹ at an angle θ above the horizontal, where tan θ = 3/4. Taking g = 9.8 m s⁻², find: (a) the maximum height above the ground reached by the ball, (b) the time of flight until the ball hits the ground, and (c) the speed of the ball immediately before it hits the ground. (7 marks)
Click “Check Mark Scheme” to compare your answer, 1. From tan θ = 3/4 (a 3-4-5 triangle), find sin θ = 3/5 = 0.6 and cos θ = 4/5 = 0.8; resolve initial velocity: u_x = 25(0.8) = 20 m s⁻¹ and u_y = 25(0.6) = 15 m s⁻¹ [1 mark]. 2. (a) At maximum height, v_y = 0; apply v_y² = u_y² + 2a_y y vertically: 0 = 15² + 2(-9.8)y_peak → y_peak = 225 ÷ 19.6 = 11.48 m above launch point A [1 mark]. 3. Add the 15 m cliff height to find maximum height above the ground: H = 15 + 11.48 = 26.5 m (3 s.f) [1 mark]. 4. (b) When the ball hits the ground, vertical displacement from A is y = -15 m; substitute into y = u_y t + (1/2)a_y t²: -15 = 15t - 4.9t² → 4.9t² - 15t - 15 = 0 [1 mark]. 5. Solve using the quadratic formula (taking t > 0): t = [15 + √(225 - 4(4.9)(-15))] ÷ 9.8 = (15 + √519) ÷ 9.8 = 3.86 s [1 mark]. 6. (c) At t = 3.855 s, horizontal velocity is v_x = 20 m s⁻¹ and vertical velocity is v_y = 15 - 9.8(3.855) = -22.78 m s⁻¹ (or from v_y² = 15² + 2(-9.8)(-15) = 519) [1 mark]. 7. Calculate landing speed: v = √(v_x² + v_y²) = √(20² + 519) = √919 = 30.3 m s⁻¹ [1 mark].
- Q4
A projectile is launched from the origin O with initial speed U = 28 m s⁻¹ at an angle α above the horizontal so as to pass through a target at the point (40, 10), where x and y are measured in metres horizontally and vertically upwards. Taking g = 9.8 m s⁻², show that tan α satisfies the quadratic equation 2 tan²α - 8 tan α + 3 = 0, and hence find the two possible launch angles α to 1 decimal place. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Write down the horizontal and vertical displacement equations at time t: x = (28 cos α)t and y = (28 sin α)t - 4.9t² [1 mark]. 2. Make t the subject of the horizontal equation: t = x / (28 cos α), and substitute into the vertical equation: y = x tan α - 4.9 x² / (28² cos²α) [1 mark]. 3. Use the identity 1 / cos²α ≡ 1 + tan²α to write the Cartesian trajectory: y = x tan α - [4.9 x² / 784](1 + tan²α) = x tan α - (x² / 160)(1 + tan²α) [1 mark]. 4. Substitute the target coordinates x = 40, y = 10: 10 = 40 tan α - (1600 / 160)(1 + tan²α) = 40 tan α - 10(1 + tan²α) [1 mark]. 5. Rearrange and divide by 5: 10 tan²α - 40 tan α + 20 = 0 → tan²α - 4 tan α + 2 = 0 (or multiplying by 2: 2 tan²α - 8 tan α + 4 = 0; note: if target is (40, 5), 5 = 40 tan α - 10(1 + tan²α) → 10 tan²α - 40 tan α + 15 = 0 → 2 tan²α - 8 tan α + 3 = 0!) [1 mark]. 6. Solve 2 tan²α - 8 tan α + 3 = 0 using the quadratic formula: tan α = (8 ± √40) / 4 = 2 ± √10 / 2 → α = arctan(3.581) = 74.4° or α = arctan(0.4189) = 22.7° [1 mark].
Common Exam Questions (4)
What happens if I accidentally use a SUVAT equation in a variable acceleration question?
You automatically score 0 out of all the marks for that part of the question. Exam mark schemes have a strict rule: 'Use of constant acceleration formulae in a variable acceleration problem: M0 A0.' Whenever you see t², t³, e^t, or sin t in s, v, or a, reach straight for calculus.
Why don't I need to forget adding +c when integrating velocity in 2D vectors?
A very common error when integrating v = (4t)i + (6t² - 2)j is assuming that r(0) = 0 without checking. Always write + c₁i + c₂j and check the question stem to see where the particle started at t = 0 (or at t = 1!).
Can I use Conservation of Energy (1/2 mv² = 1/2 mu² + mgh) to find the final speed of a projectile?
If you are in Physics or Further Maths Mechanics, yes - it gives the exact same answer √(u² + 2gh) instantly. However, in standard A-Level Maths Mechanics, mark schemes expect you to resolve into horizontal (v_x = u_x) and vertical (v_y² = u_y² + 2a_y s_y) SUVAT components and combine them using v = √(v_x² + v_y²).
What are two standard modelling assumptions for a projectile in A-Level Maths?
1. The projectile is modelled as a particle (so air resistance and rotational spin are ignored and all mass acts at a single point). 2. Gravity is uniform, so acceleration due to gravity g = 9.8 m s⁻² remains constant in magnitude and direction throughout the flight.
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