One Picture for a Whole Journey
A hotel lift leaves the ground floor, speeds up for $5\ \text{s}$, travels at a steady speed for $5\ \text{s}$, then slows down for $5\ \text{s}$ before stopping at the roof garden. The displacement–time graph below records the whole ride. Drag the time slider: a marker slides along the curve, and the steepness of the curve under the marker is the lift's velocity at that instant.
displacement $s$ = 9.00 m · gradient = velocity $v$ = 2.00 m s-1 · stage: steady speed
The amber dashed line is the tangent at the marker. Its steepness is the gradient — the velocity. Watch the gradient grow while the lift speeds up, stay constant in the middle, and fall to zero as it stops.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. The displacement of a moving body is measured from:
R2. The velocity of a body is the rate of change of its:
R3. A trapezium has parallel sides $a$ and $b$, with perpendicular distance $h$ between them. Its area is:
On a displacement–time graph the height is the displacement and the steepness is the velocity. Steeper means faster; the sign of the slope gives the direction of motion.
| Feature of the graph | What it tells you |
|---|---|
| gradient | the velocity — steeper means faster |
| horizontal line | at rest: velocity zero |
| negative gradient | moving in the negative direction |
| curve | the velocity is changing, so the body is accelerating |
Mini-example from the lift: the displacement rises from $5\ \text{m}$ at $t = 5\ \text{s}$ to $15\ \text{m}$ at $t = 10\ \text{s}$, so the gradient there — and the velocity — is $\dfrac{15 - 5}{10 - 5} = 2\ \text{m s}^{-1}$.
On a velocity–time graph the roles swap: the height is now the velocity, the gradient is the acceleration, and the area between the line and the time axis is the displacement.
| Feature of the graph | What it tells you |
|---|---|
| gradient | the acceleration |
| area between the line and the time axis | the displacement — area below the axis counts negative |
| horizontal line | constant velocity, so the acceleration is zero |
| crossing the time axis | the body reverses its direction of motion |
A straight velocity–time line means constant acceleration, so its area is a trapezium: $\dfrac{1}{2}(u + v)t$.
Adding the areas with their signs gives the displacement. Adding the sizes of the areas gives the total distance travelled. The two are equal only while the body never turns round.
The speed rises from $0$ to $20\ \text{m s}^{-1}$ in $10\ \text{s}$, stays at $20\ \text{m s}^{-1}$ for $15\ \text{s}$, then falls to zero in $5\ \text{s}$.
Areas: $\dfrac{1}{2} \times 10 \times 20 = 100\ \text{m}$, then $15 \times 20 = 300\ \text{m}$, then $\dfrac{1}{2} \times 5 \times 20 = 50\ \text{m}$.
Displacement $= 100 + 300 + 50 = 450\ \text{m}$, so the average velocity is $\dfrac{450}{30} = 15\ \text{m s}^{-1}$ — the trapezium area divided by the total time, not the average of the three speeds.
Groups of 4: tap a card on the left, then its match on the right to draw a colour-coded line. The six left-hand cards are features of motion graphs; the right-hand cards are what those features tell you. When all six lines are drawn, press Check to score. Discuss any red links together, then press Reset to try the shuffle again.
The straight line is a velocity–time graph for motion with constant acceleration. Set the initial velocity u and the acceleration a, and the shaded area shows the displacement over the first 6 s. Your task: adjust the two sliders until the shaded area equals the target 36 m — the badge fires when you are close enough. Watch how the rectangle and triangle parts of the area respond to each slider, then answer the three challenges.
A car travels along a street. It starts from rest and its speed increases uniformly at $2\ \text{m s}^{-2}$ for $10\ \text{s}$. The car then travels at a constant speed for $15\ \text{s}$, before slowing down uniformly and coming to rest after a further $5\ \text{s}$. Find (i) the speed reached after the first $10\ \text{s}$, (ii) the total distance travelled in the $30\ \text{s}$ and (iii) its average speed.
Sketch the speed–time graph first: a straight rise, a horizontal line, then a straight fall. The distance travelled is the area underneath, and the average speed is that area divided by the total time.
The speed rises steadily from $0$ to $v = 0 + 2 \times 10 = 20\ \text{m s}^{-1}$, stays level at $20\ \text{m s}^{-1}$ for $15\ \text{s}$, then falls to zero over $5\ \text{s}$.
$\dfrac{1}{2} \times 10 \times 20 + 15 \times 20 + \dfrac{1}{2} \times 5 \times 20 = 100 + 300 + 50 = 450\ \text{m}$.
$\dfrac{450}{30} = 15\ \text{m s}^{-1}$. This is the whole trapezium area divided by its width: $15\ \text{m s}^{-1}$ is the height halfway along, not the average of $0$, $20$ and $20$.
The displacement–time graph for a car's journey lasting $600\ \text{s}$ consists of two curved parts AB and CD joined by a straight section BC. The straight line BC is a tangent to both curves: at B ($t = 80$) and at C ($t = 560$). The gradient of the graph is zero at $t = 0$ and at $t = 600$, and the acceleration is constant during $0 < t < 80$ and during $560 < t < 600$. The displacement of the car is $400\ \text{m}$ when $t = 80$. Find (i) the velocity at $t = 80$, (ii) the total distance of the journey and (iii) the acceleration of the car for $0 < t < 80$.
Sketch the velocity–time graph. BC being tangent at both ends means the velocity is the same at B and at C, so the middle stage is a rectangle. Over the curved stages the velocity changes at a constant rate, so the average velocity is the mean of the two end values.
Over the first $80\ \text{s}$ the average velocity is $\dfrac{400}{80} = 5\ \text{m s}^{-1}$. With constant acceleration this is the mean of the start and end velocities, and the car starts from rest (the gradient at $t = 0$ is zero), so $\dfrac{0 + v}{2} = 5$ gives $v = 10\ \text{m s}^{-1}$ at $t = 80$.
BC is tangent to both curves, so the velocity does not change across it: the car moves at $10\ \text{m s}^{-1}$ for $560 - 80 = 480\ \text{s}$. Distance $= 480 \times 10 = 4800\ \text{m}$.
AB: average velocity $\dfrac{0 + 10}{2} = 5\ \text{m s}^{-1}$ over $80\ \text{s}$, giving $400\ \text{m}$ as stated. CD: average velocity $5\ \text{m s}^{-1}$ over $600 - 560 = 40\ \text{s}$, giving $200\ \text{m}$. Total $= 400 + 4800 + 200 = 5400\ \text{m}$.
$a = \dfrac{10 - 0}{80} = 0.125\ \text{m s}^{-2}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. The gradient of a displacement–time graph represents:
Q2. The area between a velocity–time graph and the time axis represents:
Q3. A horizontal line on a velocity–time graph means that the body:
Q4. A velocity–time graph crosses the time axis. At the crossing:
Work each on paper, then reveal the solution to check your working.
A person leaves home at 9.00 am and walks to a bus stop, covering $1.5\ \text{km}$ in $18\ \text{minutes}$. She then waits $6\ \text{minutes}$ for the bus, which takes her $9\ \text{km}$ into town in a further $6\ \text{minutes}$. (i) What is happening during the wait? (ii) The bus section is much steeper than the walking section of the distance–time graph — what does this tell you? (iii) Sketch the speed–time graph for the journey. (iv) State one simplification made in drawing these graphs.
Reveal Solution
(i) The distance from home does not change, so she is stationary: her speed is zero. (ii) Steeper means faster: the walking speed is $1.5 \div 0.3 = 5\ \text{km h}^{-1}$ while the bus travels $9\ \text{km}$ in $0.1\ \text{h}$, a speed of $90\ \text{km h}^{-1}$. (iii) The speed–time graph has three horizontal steps: $5\ \text{km h}^{-1}$ for $18\ \text{min}$, zero for $6\ \text{min}$, then $90\ \text{km h}^{-1}$ for $6\ \text{min}$. (iv) Each stage is drawn at a steady speed, with no speeding up or braking (and the walker turns round instantly).
The diagram is a sketch of the position–time graph for a fairground ride. The position rises steadily from O at ground level to a maximum at A, remains constant from A to B, then falls steadily back to ground level at C. (i) Describe what the rider is doing during OA, AB and BC. (ii) State the velocity of the ride during AB. (iii) Suggest a fairground ride that fits this graph.
Reveal Solution
(i) OA: the rider rises at a steady speed. AB: the rider is held at the top, stationary. BC: the rider comes back down at a steady speed. (ii) The graph is horizontal, so the velocity is $0$. (iii) A tower or drop ride that lifts passengers up, holds them at the top and then returns them to the ground.
A cyclist sets off from rest: her speed increases uniformly at $3\ \text{m s}^{-2}$ for $5\ \text{s}$. She then rides at a constant speed for $10\ \text{s}$ before braking uniformly, coming to rest in a further $10\ \text{s}$. (i) Calculate the acceleration for each part of the journey. (ii) Sketch an acceleration–time graph. (iii) Find the total distance travelled.
Reveal Solution
(i) First part $3\ \text{m s}^{-2}$; middle part $0$; last part the speed falls from $15\ \text{m s}^{-1}$ to zero in $10\ \text{s}$, so $a = -1.5\ \text{m s}^{-2}$. (ii) The acceleration–time graph is three horizontal steps: $3$, $0$, then $-1.5\ \text{m s}^{-2}$. (iii) Areas of the speed–time graph: $\dfrac{1}{2} \times 5 \times 15 + 10 \times 15 + \dfrac{1}{2} \times 10 \times 15 = 37.5 + 150 + 75 = 262.5\ \text{m}$.
For each situation, sketch a graph of position against time. Take the origin at the starting point, with the positive direction downwards. (i) A stone is dropped from a bridge $40\ \text{m}$ above a river. (ii) A parachutist jumps from a helicopter hovering at $2000\ \text{m}$ and opens her parachute after $10\ \text{s}$ of free fall. (iii) A bungee jumper jumps from a high bridge on an elastic cord.
Reveal Solution
(i) A curve starting with zero gradient (the stone is released from rest) that bends steeper and steeper, reaching position $40\ \text{m}$ (downwards) when it hits the river. (ii) For the first $10\ \text{s}$ the gradient grows steadily — falling faster and faster. When the parachute opens the gradient drops sharply to a small, nearly constant value, so the rest of the graph is close to a straight line of gentle slope running down to $2000\ \text{m}$. (iii) During the fall the gradient increases; as the cord stretches the gradient falls, flattening briefly at the lowest point, then the jumper rises again and smaller bounces follow — a curve with a deep dip and diminishing ripples.
A train was scheduled to travel at $50\ \text{m s}^{-1}$ for $15\ \text{minutes}$, but signals forced it to stop. Its velocity–time graph has these stages: AB level at $50\ \text{m s}^{-1}$, BC sloping straight down to the time axis, CD lying along the time axis, DE sloping straight back up to $50\ \text{m s}^{-1}$, then EF level again. Without any calculations, describe what was happening to the train during BC, CD and DE, and explain how the graph shows that it stopped.
Reveal Solution
BC: the train is slowing down at a steady rate — braking towards a standstill. CD: the velocity is zero, so the train is at rest, waiting for the signal. DE: the train accelerates steadily from rest back up to $50\ \text{m s}^{-1}$. The graph touches the time axis at C and stays there until D, so the train was really stopped — this is where the delay builds up.
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