Motion Worked ExamplesIGCSE Physics 0625

(a) Convert to SI units first: $2.4\text{ km} = 2400\text{ m}$ and $10\text{ min} = 600\text{ s}$.

(b) Average speed uses the total distance and the total time — and the time spent stopped still counts.

Total time: $10 + 3 + 7 = 20\text{ min}$.

Divide.

(a) Read the final distance from the vertical axis. Each small square is $50\text{ m}$, so the line ends at $900\text{ m}$.

(b) AB (CD is also correct). The line is horizontal — the distance is not changing, so the walker is stationary.

(c) Speed is the gradient of a distance–time graph. Read the change in distance and time across BC.

(d) Average speed over the whole graph.

With no air resistance, the only force on the stone is its weight, which stays the same. A constant force on a constant mass gives a constant acceleration — the acceleration of free fall, $g \approx 9.8\text{ m/s}^2$.

A constant acceleration means the speed increases by the same amount ($9.8\text{ m/s}$) every second. So the speed increases and the acceleration is constant: A.

B confuses "speed increasing" with "acceleration increasing". C describes something at terminal velocity (or at rest), with zero acceleration. D describes a fall with air resistance, where the growing drag reduces the resultant force.

(a) Acceleration is the change in velocity per unit time (1).

(b) Show each conversion separately: $81\text{ km} = 81\,000\text{ m}$ (1) and $1\text{ h} = 3600\text{ s}$ (1).

(c) Start from $a = \dfrac{\Delta v}{\Delta t}$ and rearrange for time.

(d) For uniform acceleration from rest, the speed–time graph is a straight line from the origin, so the distance is the area of a triangle.

(a) Marking points: the tram accelerates, then moves at constant speed, then decelerates to rest (1); the acceleration and deceleration are both constant (uniform), because those sections are straight lines (1).

(b) Distance is the area under the speed–time graph. Split it into a triangle, a rectangle and a triangle.

Add the areas (or use the trapezium: $\tfrac{1}{2}(110 + 60) \times 15$).

(c) Acceleration is the gradient of the speed–time graph. The speed falls from $15\text{ m/s}$ to $0$ between $80\text{ s}$ and $110\text{ s}$.

The negative sign shows a deceleration of $0.50\text{ m/s}^2$.

(d) Any one change in the motion: a smaller acceleration; a smaller deceleration; a lower constant (maximum) speed.

(a) The speed stays at $8.0\text{ m/s}$ until the brakes are applied at $0.6\text{ s}$ (six small squares of $0.1\text{ s}$). Reaction time $= 0.6\text{ s}$.

(b) Area of the rectangle (distance while reacting).

Area of the triangle (distance while braking). Braking lasts $3.1 - 0.6 = 2.5\text{ s}$.

Add the two areas.

(c) Use the gradient of the braking section. The velocity decreases, so the acceleration is negative.

Acceleration is the gradient of a speed–time graph. A changing acceleration means a changing gradient — a curved section.

P is straight and rising: constant acceleration. R is horizontal: zero acceleration (constant speed). S is straight and falling: constant deceleration. Only Q is curved, with a gradient that gets smaller — the acceleration decreases.

So the answer is A. B picks every section where the car speeds up; C and D treat a deceleration or a non-horizontal line as "changing", but a straight line has a constant gradient.

Draw a ruled straight line that touches the curve at $t = 6.0\text{ s}$ and has the same slope there — equal gaps between line and curve on either side (1).

Use a large triangle. The tangent crosses the speed axis at $7.0\text{ m/s}$ (at $t = 0$) and the time axis at $t = 16\text{ s}$ (1).

Evaluate. The gradient is negative, so the boat is decelerating (1).

(a) On a distance–time graph the gradient is the speed. OA gets steeper, so the speed increases: accelerating (1). BC gets less steep, so the speed decreases: decelerating (1).

(b) The maximum speed is where the graph is steepest — the straight section AB. Find its gradient.

Convert to km/h: in one hour ($3600\text{ s}$) she would cover $8.0 \times 3600 = 28\,800\text{ m} = 28.8\text{ km}$.

(c) Average speed is total distance over total time, including the rest in CD.

(a) Find the time for each $10\text{ cm}$ interval by subtracting successive times: $0.28$, $0.18$, $0.14$, $0.12$, $0.12$, $0.12\text{ s}$.

After band $4$, every interval takes the same time ($0.12\text{ s}$): equal distances in equal times means constant speed — terminal velocity (1 band, 1 reason).

(b) Use one of the constant intervals, or better, all three together ($0.300\text{ m}$ in $0.96 - 0.60 = 0.36\text{ s}$).

(c) Marking points: at release the ball's speed is zero, so there is no resistance and it accelerates downwards (1). As the speed increases, the resistance of the oil increases, so the resultant force and the acceleration decrease (1). When the upward forces balance the weight, the resultant force is zero, and the ball falls at a constant speed, its terminal velocity (1).

(a) Rearrange $a = \dfrac{\Delta v}{\Delta t}$.

(b) While the cyclist is faster, the gap grows; once the car is faster, it shrinks. So the gap is greatest when their speeds are equal, at $12\text{ m/s}$.

Distances at $8.0\text{ s}$ are areas under each speed–time line: a rectangle for the cyclist, a triangle for the car.

(c) At $t = 12\text{ s}$: cyclist $12 \times 12 = 144\text{ m}$; car $\tfrac{1}{2} \times 12 \times 18 = 108\text{ m}$. The car is still behind.

After $12\text{ s}$ the car gains $18 - 12 = 6\text{ m}$ every second.

Distance from the lights (check with both).

(a) In free fall the acceleration is $g$. Rearrange $a = \dfrac{\Delta v}{\Delta t}$.

(b) The speed–time graph is a straight line from the origin to $(1.5\text{ s},\ 14.7\text{ m/s})$. The area is a triangle.

(c) On the planet, after a time $t$ the speed is $3.7t$. The area of the triangle under its graph must equal the same $11.025\text{ m}$.

Take the square root.

(d) The gravitational field strength is smaller on the planet, so the weight (the gravitational force on each kilogram) is smaller (1). So the acceleration of the ball is smaller, and it takes longer to reach the ground (1).