Distance-Time & Velocity-Time Graphs, Acceleration & Terminal Velocity
Motion in a straight line is analysed using vector quantities (displacement, velocity, and acceleration) that have both magnitude and direction, alongside scalar quantities (distance and speed) that have magnitude only. On a distance-time graph, the gradient represents speed, whereas on a velocity-time graph, the gradient represents acceleration (a = Δv / t) and the area enclosed under the graph represents the total distance travelled (displacement). Examiners frequently test multi-stage velocity-time graph area calculations, the uniform acceleration equation v² - u² = 2as, and the 6-mark explanation of terminal velocity using Newton's First and Second Laws.
- Scalars vs Vectors (Speed vs Velocity)
- Distance and speed are scalars (magnitude only, e.g., 20 m/s). Displacement and velocity are vectors (magnitude and direction, e.g., 20 m/s due north). Formula: Distance travelled, s (m) = Speed, v (m/s) x Time, t (s). Note: If an object moves in a circle at a constant speed (e.g., a satellite or a car on a roundabout), its velocity is constantly changing because its direction of motion is constantly changing, meaning it is accelerating.
- Distance-Time (d-t) Graphs
- Distance is plotted on the y-axis and time on the x-axis. 1. Gradient = Vertical change / Horizontal change = Speed (m/s). 2. Horizontal flat line = stationary (speed = 0 m/s). 3. Straight sloping line = constant speed (steeper slope = higher speed). 4. Upward-curving line (increasing gradient) = accelerating; find instantaneous speed at a point on a curve by drawing a tangent to the curve and calculating the gradient of the tangent.
- Acceleration Equations (a = Δv / t and v² - u² = 2as)
- 1. Acceleration, a (m/s²) = Change in velocity, Δv (m/s) / Time taken, t (s), where Δv = final velocity (v) - initial velocity (u). If an object slows down, v < u so acceleration is negative (deceleration). 2. Uniform acceleration equation (on physics equation sheet): (Final velocity)² - (Initial velocity)² = 2 x Acceleration x Distance -> v² - u² = 2as. Near the Earth's surface, an object falling freely under gravity alone has a uniform acceleration of g = 9.8 m/s².
Distance-Time & Velocity-Time Graphs, Acceleration & Terminal Velocity: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Scalars vs Vectors (Speed vs Velocity)
- Distance and speed are scalars (magnitude only, e.g., 20 m/s). Displacement and velocity are vectors (magnitude and direction, e.g., 20 m/s due north). Formula: Distance travelled, s (m) = Speed, v (m/s) x Time, t (s). Note: If an object moves in a circle at a constant speed (e.g., a satellite or a car on a roundabout), its velocity is constantly changing because its direction of motion is constantly changing, meaning it is accelerating.
- Distance-Time (d-t) Graphs
- Distance is plotted on the y-axis and time on the x-axis. 1. Gradient = Vertical change / Horizontal change = Speed (m/s). 2. Horizontal flat line = stationary (speed = 0 m/s). 3. Straight sloping line = constant speed (steeper slope = higher speed). 4. Upward-curving line (increasing gradient) = accelerating; find instantaneous speed at a point on a curve by drawing a tangent to the curve and calculating the gradient of the tangent.
- Acceleration Equations (a = Δv / t and v² - u² = 2as)
- 1. Acceleration, a (m/s²) = Change in velocity, Δv (m/s) / Time taken, t (s), where Δv = final velocity (v) - initial velocity (u). If an object slows down, v < u so acceleration is negative (deceleration). 2. Uniform acceleration equation (on physics equation sheet): (Final velocity)² - (Initial velocity)² = 2 x Acceleration x Distance -> v² - u² = 2as. Near the Earth's surface, an object falling freely under gravity alone has a uniform acceleration of g = 9.8 m/s².
- Velocity-Time (v-t) Graphs
- Velocity is plotted on the y-axis and time on the x-axis. 1. Gradient of the line = Acceleration (m/s²). 2. Horizontal flat line = constant velocity (zero acceleration). 3. Straight upward slope = uniform acceleration; straight downward slope = uniform deceleration. 4. Area enclosed under the velocity-time graph = Total distance travelled (displacement, in metres) (Higher Tier).
- Terminal Velocity (Falling through a Fluid)
- When an object falls through a fluid (air or liquid): 1. Initially, downward Weight (W = mg) > upward Drag (air resistance), so there is a large downward resultant force and large acceleration (initially 9.8 m/s²). 2. As velocity increases, upward drag force increases (more frequent/harder collisions with air molecules), decreasing the downward resultant force and decreasing acceleration. 3. Eventually, upward Drag = downward Weight; resultant force = 0 N, so acceleration = 0 m/s² and the object falls at a constant terminal velocity.
- Required Practical: Investigating F = ma (Trolley on a Ramp)
- Use a dynamics trolley on a runway pulled by slotted masses hanging over a pulley on a string, passing through two light gates connected to a data logger to measure acceleration. 1. To investigate acceleration vs Force (at constant total mass): transfer slotted masses from the top of the trolley onto the hanging hook (so the total mass of the accelerating system m_trolley + m_hanging stays constant). 2. To investigate acceleration vs Mass (at constant force): keep the hanging weight constant and add extra masses to the top of the trolley.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Why is a satellite orbiting the Earth at a constant speed of 7,800 m/s accelerating?
What does the gradient represent on: (a) a distance-time graph, and (b) a velocity-time graph?
How do you find the instantaneous acceleration at t = 4 s on a curved velocity-time graph, and how do you find the distance travelled?
State the two equations for acceleration in GCSE Physics.
In the F = ma required practical, why must you move masses from the trolley onto the weight hanger when varying the pulling force?
What is the resultant force on a skydiver falling at a constant terminal velocity of 55 m/s?
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 train starts from rest at a station and accelerates uniformly at 1.5 m/s² for 20 s. It then travels at a constant velocity for 60 s, before decelerating uniformly to a stop in 10 s at the next station. Calculate: (a) the maximum velocity reached by the train, (b) the deceleration in the final 10 s, and (c) the total distance between the two stations. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. (a) Maximum velocity after 20 s: v = u + at = 0 + (1.5 x 20) = 30 m/s [1 mark]. 2. (b) Deceleration in final 10 s: a = (v - u) / t = (0 - 30) / 10 = -3.0 m/s² (or deceleration of 3.0 m/s²) [1 mark]. 3. (c) Total distance = total area under the velocity-time graph (triangle 1 + rectangle + triangle 2) [1 mark]. 4. Area = (½ x 20 x 30) + (60 x 30) + (½ x 10 x 30) = 300 m + 1,800 m + 150 m [1 mark]. 5. Total distance = 2,250 m (or 2.25 km) [1 mark].
- Q2
Explain in terms of forces and motion how a skydiver reaches terminal velocity after jumping from an aeroplane, and explain what happens to the forces and motion when they subsequently open their parachute. [6 marks]
Click “Check Mark Scheme” to compare your answer, 1. Immediately after jumping, the skydiver's downward velocity is zero so upward air resistance is zero; the only vertical force is Weight acting downwards, causing the skydiver to accelerate downwards at 9.8 m/s² [1 mark]. 2. As the skydiver's downward speed increases, they collide more frequently with air molecules, so upward air resistance increases [1 mark]. 3. This reduces the downward resultant force (Weight: Air resistance), which decreases the skydiver's acceleration (by F = ma), so the velocity-time graph curves and becomes less steep [1 mark]. 4. Eventually, upward air resistance equals downward Weight; the resultant force is zero, acceleration is zero, and the skydiver falls at a constant high terminal velocity [1 mark]. 5. When the parachute opens, the surface area increases dramatically, causing upward air resistance to become much greater than downward Weight [1 mark]. 6. There is now a large upward resultant force which causes the skydiver to decelerate (slow down rapidly while still moving downwards), reducing air resistance until it once again equals Weight at a much lower, safe terminal velocity for landing [1 mark].
- Q3
A aircraft lands on a runway at an initial velocity of 70 m/s and decelerates uniformly at -2.5 m/s² until its velocity reaches 10 m/s. Calculate the distance travelled by the aircraft along the runway during this deceleration. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Select the uniform acceleration equation: v² - u² = 2 x a x s [1 mark]. 2. Substitute given values (u = 70 m/s, v = 10 m/s, a = -2.5 m/s²): (10)² - (70)² = 2 x (-2.5) x s -> 100 - 4,900 = -5.0 x s [1 mark]. 3. Rearrange and solve for distance s: s = -4,800 / -5.0 = 960 m [1 mark].
- Q4
Describe how a student could use a trolley, a runway, a pulley, string, slotted masses, and light gates to investigate how the acceleration of an object depends on the resultant force acting on it. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Set up a runway on a bench (tilted slightly to compensate for friction) with a pulley at the end; attach a trolley to a string passing over the pulley to a hook of slotted masses, and place five 100 g slotted masses on top of the trolley [1 mark]. 2. Measure the total mass of the system (trolley + all slotted masses) using a balance, and position two light gates along the runway connected to a data logger to measure acceleration directly [1 mark]. 3. Release the trolley from rest, record the pulling force (weight of the hanging hook: F = mg) and the acceleration from the light gates, and repeat 3 times to calculate a mean acceleration [1 mark]. 4. Transfer one 100 g slotted mass from the top of the trolley onto the hanging hook (this increases the pulling force F by ~0.98 N while keeping the total mass of the accelerating system constant) [1 mark]. 5. Repeat for 5 different values of pulling force and plot a graph of acceleration (y-axis) against resultant force (x-axis), which should be a straight line through the origin (a ∝ F) [1 mark].
Common Exam Questions (4)
How do you avoid confusing distance-time graphs with velocity-time graphs?
Always look at the y-axis label first. 1. A horizontal flat line on a distance-time graph means the object is stationary (stopped, speed = 0 m/s), whereas a horizontal flat line on a velocity-time graph means the object is moving at a constant velocity (acceleration = 0 m/s²). 2. A straight upward-sloping diagonal line on a distance-time graph means constant speed, whereas on a velocity-time graph it means constant acceleration.
How do you find the distance travelled from a velocity-time graph?
The distance travelled (or displacement) in any time interval is equal to the area under the velocity-time graph. Split the shape under the line into simple right-angled triangles (Area = ½ x base x height) and rectangles (Area = base x height), or trapeziums, and add the areas together. If the line is curved (Higher Tier), count the grid squares under the curve and multiply by the distance represented by one square.
Why does a skydiver reach terminal velocity? (4-mark chain of reasoning)
1. At the instant of jumping, speed is zero so air resistance (drag) is zero; the only force is weight acting downwards, giving an initial downward acceleration of g = 9.8 m/s². 2. As the skydiver falls faster, air resistance acting upwards increases, which reduces the downward resultant force and therefore decreases the acceleration (by F = ma). 3. Eventually, upward air resistance grows until it equals the skydiver's downward weight. 4. At this point, the resultant force is zero (0 N), so acceleration is zero and the skydiver falls at a constant maximum velocity called terminal velocity.
Does a skydiver move upwards when they open their parachute?
No. This is a classic physics misconception. When the parachute opens, surface area increases hugely, so upward air resistance suddenly becomes much larger than downward weight. The resultant force acts upwards, which causes the skydiver to decelerate (slow down rapidly) while still falling downwards, until air resistance drops back down to equal weight at a new, safe low terminal velocity.
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