Fig. 1.1 shows the speed-time graph for a car travelling along a horizontal road. (a) On Fig. 1.1, mark and label a point where the car has a non-uniform deceleration. (b) Calculate the deceleration of the car at $t = 3.0\ \text{s}$. deceleration = \_\_\_\_\_\_ (c) Explain, in terms of the horizontal forces that act on the car, why its speed is constant at $t = 1.0\ \text{s}$.

(a) Point marked on the curved deceleration section between t = 4.0 s and t = 8.0 s · (b) 2.5 m/s^2 · (c) The forward driving force and backward resistive forces are equal, so the resultant horizontal force is zero.

Physics 5054/22 — May/June 2016

Cambridge O-Level · Theory · worked solutions for every part, with the mark scheme

questions

marks

105

minutes

Topics Forces · Energy, Work and Power · Current, Voltage and Resistance · [Legacy] Introductory Electronics · Kinematics · Mass, Weight and Density · +13 more

Q1KinematicsForcesFree sample

Fig. 1.1 shows the speed-time graph for a car travelling along a horizontal road.

Fig. 1.1

(a)

On Fig. 1.1, mark and label a point where the car has a non-uniform deceleration.

DifficultyEasy

Worked solution

Answer

Mark a point on the curved section of the graph between $t = 4.0$ s and $t = 8.0$ s (for example, at $t = 5.0$ s, speed $= 24$ m/s). This section has a changing gradient, indicating non-uniform deceleration.

Final answer

Point marked on the curved deceleration section between t = 4.0 s and t = 8.0 s

Detailed explanation

Walkthrough

The gradient of a speed-time graph gives the acceleration. A straight line has a constant gradient, meaning uniform acceleration (or deceleration). A curved line has a changing gradient, meaning non-uniform acceleration. The graph shows a straight-line deceleration from $t = 2.0$ s to $t = 4.0$ s, followed by a curved section from $t = 4.0$ s to $t = 8.0$ s where the speed decreases from 25 m/s to 22.5 m/s. Any point on this curved section has non-uniform deceleration.

Key Takeaways

A straight section on a speed-time graph represents uniform acceleration; a curved section represents non-uniform acceleration.

Common Mistakes

Marking a point on the straight-line section between $t = 2.0$ s and $t = 4.0$ s (this is uniform deceleration) or on the horizontal sections (this is zero acceleration, not deceleration).

Things to Be Careful About

Ensure the marked point is clearly on the curve and not exactly on a corner where the straight line meets the curve, as the exact boundary can be ambiguous. Mark it clearly in the middle of the curved region, e.g., between $t = 4.0$ s and $t = 7.5$ s.

Techniques used

identify non-uniform acceleration from a speed-time graph

(b)

Calculate the deceleration of the car at $t = 3.0\ \text{s}$ .

deceleration = ______

DifficultyMedium-Easy

Worked solution

Working

At $t = 3.0$ s, the graph is a straight line from $(2.0, 30)$ to $(4.0, 25)$ . The deceleration is the magnitude of the gradient of this line.

\text{deceleration} = \frac{v - u}{t} = \frac{30 - 25}{4.0 - 2.0} = \frac{5}{2.0} = 2.5 \text{ m/s}^2

Answer

$2.5 \text{ m/s}^2$

Final answer

2.5 m/s^2

Detailed explanation

Walkthrough

The question asks for the deceleration at $t = 3.0$ s. At this time, the car is on the straight-line section of the graph between $t = 2.0$ s and $t = 4.0$ s. For a straight line on a speed-time graph, the acceleration is constant and equals the gradient. We read the coordinates at the ends of this straight section: $(u, t_1) = (30 \text{ m/s}, 2.0 \text{ s})$ and $(v, t_2) = (25 \text{ m/s}, 4.0 \text{ s})$ . The change in speed is $25 - 30 = -5$ m/s over a time of $4.0 - 2.0 = 2.0$ s. The acceleration is $-5 / 2.0 = -2.5 \text{ m/s}^2$ . Deceleration is the magnitude of this, so $2.5 \text{ m/s}^2$ .

Key Takeaways

The gradient of a speed-time graph gives acceleration. For a straight line, the gradient is constant and can be found using any two points on that line.

Common Mistakes

Using coordinates from different sections of the graph (e.g., mixing the straight line and the curve) or forgetting that deceleration is a positive magnitude.

Things to Be Careful About

Read the graph to the correct precision. The grid lines are every 2 s and 5 m/s, with minor divisions every 0.2 m/s and 0.4 s. Ensure the points $(2.0, 30)$ and $(4.0, 25)$ are read accurately.

Techniques used

calculate gradient of a straight-line section on a speed-time graph

(c)

Explain, in terms of the horizontal forces that act on the car, why its speed is constant at $t = 1.0\ \text{s}$ .

DifficultyMedium-Easy

Worked solution

Answer

The forward driving force and the backward resistive forces (friction and air resistance) are equal in magnitude. Therefore, the resultant horizontal force is zero, so there is no acceleration and the speed remains constant.

Final answer

The forward driving force and backward resistive forces are equal, so the resultant horizontal force is zero.

Detailed explanation

Walkthrough

At $t = 1.0$ s, the speed-time graph is horizontal, meaning the speed is constant at 30 m/s. Constant speed implies zero acceleration. According to Newton's first law (or $F = ma$ with $a = 0$ ), if the acceleration is zero, the resultant force must be zero. In the horizontal direction, the car experiences a forward driving force from the engine and backward resistive forces from friction and air resistance. For the resultant force to be zero, these two forces must be equal in magnitude and opposite in direction, so they balance each other.

Key Takeaways

Constant speed (or constant velocity) means zero acceleration, which in turn means the resultant force is zero. Forces must be balanced.

Common Mistakes

Saying 'there are no forces acting on the car' (forces are always present; they just cancel out) or saying 'the forces are zero' instead of 'the resultant force is zero'.

Things to Be Careful About

The question specifically asks to explain 'in terms of the horizontal forces'. You must name the forward and backward forces and state that they are equal or balance. Simply saying 'resultant force is zero' without mentioning the individual forces may only earn one mark.

Techniques used

relate constant speed to zero resultant forceidentify forward and backward horizontal forces

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