What you'll learn
- How position, displacement and distance travelled describe where an object is and the route it takes.
- How velocity differs from speed.
- How acceleration describes a change in velocity.
- How to use signs, directions and SI units correctly in kinematics calculations.
What is kinematics?
Kinematics is the study of motion without considering the forces that cause it.
At this stage, you will usually model a moving object as a particle. A particle has negligible size, so its position can be represented by a single point.
Motion is often restricted to a straight line. You must choose one direction as positive; motion in the opposite direction is then negative. For vertical motion, upwards is commonly chosen as positive, but the question may use a different convention.
Choose a positive direction
A positive sign means “in the chosen positive direction”. It does not automatically mean rightwards or upwards.
Position
To describe position on a straight line, you first choose a fixed point called the origin, usually labelled OOO. Position is then measured relative to this origin.
Position
The position of a particle is its location relative to a chosen origin. It is often represented by the coordinate xxx and measured in metres.
For example:
- x=5 mx=5\text{ m}x=5 m means the particle is 5 m from the origin in the positive direction.
- x=−3 mx=-3\text{ m}x=−3 m means the particle is 3 m from the origin in the negative direction.
A negative position does not mean that anything has gone wrong. It simply places the particle on the negative side of the origin.
Displacement
Displacement describes the overall change in position. It depends only on the initial and final positions, not on the route taken.
Displacement
The displacement of a particle is its final position minus its initial position:
displacement=xfinal−xinitial.\text{displacement}=x_{\text{final}}-x_{\text{initial}}.displacement=xfinal−xinitial.Displacement is a vector quantity, so it has both magnitude and direction.
In one-dimensional motion, direction is shown using a positive or negative sign.
Calculating displacement
A particle moves from the position x=−4 mx=-4\text{ m}x=−4 m to the position x=7 mx=7\text{ m}x=7 m. Find its displacement.
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Identify the initial and final positions: xinitial=−4 mx_{\text{initial}}=-4\text{ m}xinitial=−4 m and xfinal=7 mx_{\text{final}}=7\text{ m}xfinal=7 m.
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Subtract the initial position from the final position:
displacement=xfinal−xinitial=7−(−4)=11 m.\begin{aligned} \text{displacement} &=x_{\text{final}}-x_{\text{initial}}\\ &=7-(-4)\\ &=11\text{ m}. \end{aligned}displacement=xfinal−xinitial=7−(−4)=11 m. -
The answer is positive, so the displacement is 11 m in the chosen positive direction.
Subtracting in the wrong order
Displacement is always final position minus initial position. A particle moving from x=7 mx=7\text{ m}x=7 m to x=−4 mx=-4\text{ m}x=−4 m has displacement −11 m-11\text{ m}−11 m, not 11 m.
Distance travelled
Distance travelled
Distance travelled is the total length of the route followed by a particle. It is a scalar quantity, so it has magnitude but no direction.
Distance travelled is never negative. If a particle changes direction, add the lengths of all parts of its journey.
Displacement and distance travelled are therefore not usually the same:
- Displacement compares only the starting and finishing positions.
- Distance travelled includes every part of the route.
- The magnitude of displacement can never be greater than the distance travelled.
Comparing distance and displacement
A particle starts at x=2 mx=2\text{ m}x=2 m, moves to x=10 mx=10\text{ m}x=10 m and then returns to x=6 mx=6\text{ m}x=6 m. Find its distance travelled and displacement.
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On the first part of the journey, the particle travels 10−2=8 m10-2=8\text{ m}10−2=8 m.
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On the return journey, it travels 10−6=4 m10-6=4\text{ m}10−6=4 m. Therefore,
distance travelled=8+4=12 m.\text{distance travelled}=8+4=12\text{ m}.distance travelled=8+4=12 m. -
Its displacement depends only on its initial and final positions:
displacement=6−2=4 m.\text{displacement}=6-2=4\text{ m}.displacement=6−2=4 m. -
The particle has travelled 12 m in total, but it finishes only 4 m from its starting position in the positive direction.
Returning to the start
If a particle returns to its initial position, its displacement is zero, but its distance travelled will usually be greater than zero.
Velocity
Velocity
Velocity is the rate of change of displacement. Average velocity over a time interval is
average velocity=displacementtime taken.\text{average velocity}=\frac{\text{displacement}}{\text{time taken}}.average velocity=time takendisplacement.Velocity is a vector quantity and is measured in metres per second, m s⁻¹.
The sign of velocity indicates the direction of motion:
- Positive velocity means motion in the chosen positive direction.
- Negative velocity means motion in the opposite direction.
- Zero velocity means the particle is instantaneously at rest.
Average velocity describes motion over a whole time interval. Instantaneous velocity is the velocity at one particular instant.
Finding average velocity
A particle moves from x=−5 mx=-5\text{ m}x=−5 m to x=19 mx=19\text{ m}x=19 m in 6 s. Find its average velocity.
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Calculate the displacement:
displacement=19−(−5)=24 m.\text{displacement}=19-(-5)=24\text{ m}.displacement=19−(−5)=24 m. -
Divide the displacement by the time taken:
average velocity=246=4 m s−1.\text{average velocity}=\frac{24}{6}=4\text{ m s}^{-1}.average velocity=624=4 m s−1. -
The positive sign shows that the average velocity is in the chosen positive direction.
Speed
Speed
Speed is the rate at which distance is travelled. Average speed is
average speed=distance travelledtime taken.\text{average speed}=\frac{\text{distance travelled}}{\text{time taken}}.average speed=time takendistance travelled.Speed is a scalar quantity, measured in m s⁻¹.
Speed cannot be negative. In one-dimensional motion, the speed at any instant is the magnitude of the velocity:
speed=∣velocity∣.\text{speed}=|\text{velocity}|.speed=∣velocity∣.For example, a velocity of −7 m s−1-7\text{ m s}^{-1}−7 m s−1 corresponds to a speed of 7 m s⁻¹.
Comparing average speed and average velocity
A runner travels 100 m east and then 40 m west in a total time of 20 s. Take east as positive.
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The total distance travelled is 100+40=140 m100+40=140\text{ m}100+40=140 m, so
average speed=14020=7 m s−1.\text{average speed}=\frac{140}{20}=7\text{ m s}^{-1}.average speed=20140=7 m s−1. -
The displacement is 100−40=60 m100-40=60\text{ m}100−40=60 m east, so
average velocity=6020=3 m s−1.\text{average velocity}=\frac{60}{20}=3\text{ m s}^{-1}.average velocity=2060=3 m s−1. -
Average speed and the magnitude of average velocity are different because the runner changed direction.
Confusing speed with velocity
Use distance travelled when calculating average speed, but use displacement when calculating average velocity.
Acceleration
Acceleration
Acceleration is the rate of change of velocity. Average acceleration over a time interval is
average acceleration=final velocity−initial velocitytime taken.\text{average acceleration} =\frac{\text{final velocity}-\text{initial velocity}}{\text{time taken}}.average acceleration=time takenfinal velocity−initial velocity.Acceleration is a vector quantity, measured in metres per second squared, m s⁻².
If the initial velocity is uuu, the final velocity is vvv, the acceleration is aaa and the time taken is ttt, then
a=v−ut.a=\frac{v-u}{t}.a=tv−u.A positive acceleration acts in the chosen positive direction; a negative acceleration acts in the negative direction.
Calculating acceleration
A particle's velocity changes from −3 m s−1-3\text{ m s}^{-1}−3 m s−1 to 5 m s−15\text{ m s}^{-1}5 m s−1 in 4 s. Find its average acceleration.
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Find the change in velocity:
v−u=5−(−3)=8 m s−1.v-u=5-(-3)=8\text{ m s}^{-1}.v−u=5−(−3)=8 m s−1. -
Divide by the time taken:
a=v−ut=84=2 m s−2.a=\frac{v-u}{t}=\frac{8}{4}=2\text{ m s}^{-2}.a=tv−u=48=2 m s−2. -
The positive acceleration means that the change in velocity is in the chosen positive direction.
Acceleration and slowing down
Negative acceleration does not automatically mean that a particle is slowing down. You must compare the signs of velocity and acceleration.
- If velocity and acceleration have the same sign, the speed is increasing.
- If velocity and acceleration have opposite signs, the speed is decreasing.
For example, if both velocity and acceleration are negative, the particle is moving in the negative direction and gaining speed.
Direction versus speed
Velocity tells you the direction of motion. Acceleration tells you how velocity is changing. Use their signs together to decide whether speed is increasing or decreasing.
In the exam
- State or identify the positive direction before using signed quantities.
- Keep distance and speed separate from displacement and velocity.
- Calculate changes as final value minus initial value, especially in v−uv-uv−u.
- Include appropriate SI units and use the sign of your answer to interpret direction.
Check yourself
- A particle moves from x=3 mx=3\text{ m}x=3 m to x=−8 mx=-8\text{ m}x=−8 m. What is its displacement?
- Can a particle have zero displacement but a positive distance travelled? Explain.
- If velocity is negative and acceleration is positive, is the particle's speed necessarily increasing?
