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1.1Displacement, distance, speed and velocity

Straight line motion in one dimension means that the particle that moves is constrained to move along a straight line. The position the coordinate on the number line of the particle is then given by its coordinate on the straight line (Figure 1.1). Position is a vector quantity—this is important when discussing projectile motion but, while discussing motion on a straight line, we can just express position as a positive or a negative number since here the vector can only point in one direction or the opposite.

If the line is horizontal, we use the symbol x to represent the coordinate and position. If the line is vertical, we use the symbol y. In general, for an arbitrary line, we use a generic symbol, s, for position. So, in Figure 1.1, x = 6 m, y = −4 m, s = 0 and y = 4 m. It is up to us to decide which side of zero we call positive and which side of zero we call negative; the decision is arbitrary.

Figure 1.1: The position of a particle is determined by the coordinate on the number line.

The change in position is called displacement change in position; the difference between the position of the oscillating body and the position at equilibrium, Δs = sfinal − sinitial. Displacement is a vector.

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Table 1.1 shows four different motions. Make sure you understand how to calculate the displacement and that you understand the direction of motion.

Initial position | Final position | Displacement | Direction of motion
12 m | 28m | +16 m | Towards increasing s
−6 m | −14 m | −8 m | Towards decreasing s
10 m | −5 m | −15 m | Towards decreasing s
−20 m | −15 | +5 m | Towards increasing s

Table 1.1: Four different motions.

Consider the two motions shown in Figure 1.2. In the first motion, the particle leaves its initial position at −4 m and continues to its final position at 16 m. The displacement is:

Δ s=s final −s initial =16−(−4)=20 m

The distance length of path followed travelled is the actual length of the path followed (20 m).

In the second motion, the particle leaves its initial position at 12 m, arrives at position 20 m and then comes back to its final position at 4 m.

Figure 1.2

The second motion is an example of motion with changing direction. The change in the position of this particle (its displacement) is:

Δ s=s final −s initial =4−12=−8 m

But the distance travelled by the particle (the length of the path) is 8 m in the outward trip and 16 m on the return trip, making a total distance length of path followed of 24 m. So, we must be careful to distinguish distance from displacement:

Distance is a scalar quantity but displacement is a vector quantity.

Numerically, they are different if there is a change of direction, as in Figure 1.2.

As the particle moves on the straight line its position changes. In uniform motion motion with constant velocity in equal intervals of time, the position changes by the same amount.

For uniform motion the velocity is the displacement divided by the time to achieve that displacement: v=Δ s Δ t, the rate of change of position., v, of the particle is the displacement divided by the time to achieve that displacement: v=Δ s Δ t.

The average speed total distance travelled divided by the time taken is the total distance travelled divided by the time taken.

Assume that the first motion in Figure 1.2 took 4.0 s to complete. The velocity is 20 4.0=5.0 m s−1. The average speed is the same since the distance travelled is also 20 m.

Assume the second motion took 6.0 s to complete. The velocity is −8.0 6.0=−1.3 m s−1 and the average speed is 24 6.0=4.0 m s−1. So, in uniform motion, average speed and velocity are not the same when there is change in direction.

This implies that for uniform motion:

v=s−s initial t−0

which can be re-arranged to give:

s=s initial +v t

(Notice how we use s for final position, rather than s f or sfinal, and we will write s i for sinitial for simplicity). So for motion with constant velocity:

s=s i+v t

Exam tip

This formula can only be used when the velocity is constant.

This formula gives, in uniform motion, the position s of the moving object t seconds after time zero, given that the constant velocity is v and the initial position is si.

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This means that a graph of position against time is a straight line and the graph of velocity against time is a horizontal straight line (Figure 1.3).

Figure 1.3: In uniform motion the graph of position against time is a straight line with non-zero gradient. The graph of velocity against time is a horizontal straight line.

Positive velocity means that the position s is increasing. Negative velocity means that s is decreasing. Observe that the area under the v versus t graph from t = 0 to time t is vt.

From s = s i + vt we deduce that s − s i = vt and so the area in a velocity-against-time graph is the displacement.

CHECK YOURSELF 1

An object moves from A to B at speed 15 m s−1 and returns from B to A at speed 30 m s−1.

What is the average speed for the round trip?

Worked Example 1.1

Two cyclists, A and B, start moving at the same time. The initial position of A is 0 km and her velocity is +20 km h−1. The initial position of B is 150 km away from A and he cycles at a velocity of −30 km h−1.

aDetermine the time and position at which they will meet.

bWhat is the displacement of each cyclist when they meet?

cOn another occasion the same experiment is performed but this time B starts 1 h after A. When will they meet?

Answer

aThe position of A is given by the formula s A = 0 + 20t.

The position of B is given by the formula s B = 150 − 30t.

They will meet when they are at the same position, i.e. when s A = s B. This implies:

20 t=150−30 t 50 t=150 t=3.0 h

The common position is found from either s A = 20 × 3.0 = 60 km or s B = 150 − 30 × 3.0 = 60 km.

bThe displacement of A is 60 km − 0 = 60 km. That of B is 60 km − 150 km = −90 km.

c s A = 0 + 20t as before. When t h go by, B will have been moving for only t − 1 h.

Hence s B = 150 − 30(t − 1).

They will meet when s A = s B:

20 t=150−30(t−1)50 t=180 t=3.6 h

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TEST YOUR UNDERSTANDING

1A car must be driven a distance of 120 km in 2.5 h. During the first 1.5 h the average speed was 70 km h−1. Calculate the average speed for the remainder of the journey.

2Find the constant velocity for each motion whose position–time graphs are shown in Figure 1.4.

Figure 1.4

3Draw the position–time graphs for an object moving in a straight line with velocity–time graphs as shown in Figure 1.5. The initial position is zero.

In each case, state the displacement at 4 s.

Figure 1.5

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4Two cyclists, A and B, have displacements 0 km and 70 km, respectively.

At t = 0 they begin to cycle towards each other with velocities 15 km h−1 and 20 km h−1, respectively.

At the same time, a fly that was sitting on cyclist A starts flying towards cyclist B with a velocity of 30 km h−1.

As soon as the fly reaches cyclist B it immediately turns around and flies towards cyclist A, and so on until cyclist A and cyclist B meet.

aFind the position of the two cyclists and the fly when all three meet.

bDetermine the distance travelled by the fly.