1D Motion graphsWe will only consider straight line motion (also called ‘one dimensional’ or ‘linear’ motion). For straight line motion, direction can be indicated with a ‘+’ or ‘-’ sign as long as the positive direction has been specified.

This PowerPoint show is designed to accompany (or remind you of) an oral presentation. Without the oral presentation some parts of this PowerPoint show may be meaningless.

Position-time graphsPosition / m

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On a position-time graph, the gradient gives the velocity.

This line has zero gradient, so the velocity is zero.

Position / m

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Line A has a gradient of 4/8 = 0.5 m/s.Line A is a steeper line -> higher velocity (forwards).

On a position-time graph, the gradient gives the velocity.

Line B has a gradient of 4/10 = 0.4 m/s.

Position / m

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Negative slope so negative velocity (going backwards).

Velocity = gradient = -4/10 = -0.4 m/s

Position / m

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-6After the first 5 seconds the object moves back behind the zero position.The velocity is constant and negative: v = -12 / 10 = - 1.2 m/s

Position / m

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What is the average velocity over the 8 seconds? Over the first six seconds?Over the period t = 8s to t=10s ?

V = 4/8 = 0.5 m/sV = 2 / 6 = 0.333 m/sV = 2 / 2 = 1 m/s

Position / m

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What is the average velocity over the ten seconds? Over the first four seconds?Over the period t = 4s to t=10s ?

V = 5/10 = 0.5 m/sV = 4/4 = 1 m/sV = 1/6 = 0.167 m/s

Position / m

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What is the instantaneous velocity at t= 2s?

On a position-time graph, instantaneous velocity at any time is the slope of a tangent to

the curve at that time.

Instantaneous velocity = slope of tangent = (4.3-1.7) / 4 = 0.65 m/s

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Velocity-time graphs

• Warning: you MUST pay attention to the type of graph (p-t) or (v-t) because this changes EVERYTHING.

• Use your head to extract the meaning of each graph.

Velocity-time graphsVelocity (m/s)

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On a velocity-time graph, the gradient gives the acceleration.

Here the gradient is zero, so the acceleration is zero.

Velocity (m/s)

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Acceleration = 4 / 10 = 0.4 m/s²On a velocity-time graph, the gradient gives the acceleration.

Velocity (m/s)

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In the first four seconds: acceleration = zero

In the next 10 seconds: acceleration = -4 / 10 = -0.4 m/s²

Velocity (m/s)

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-6Velocity is negative (backwards). Acceleration = zero

Velocity (m/s)

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-3 In the first four seconds: acceleration = -2 / 4 = -0.5 m/s²In the next 6 seconds: acceleration = zero

Velocity (m/s)

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Acceleration = -8 / 8 = -1 m/s²Note that the acceleration is constant here, whether the velocity is forwards, zero or backwards (negative).

Velocity (m/s)

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In the first four seconds, a = 4 / 4 = 1 m/s²

Between t= 4s and t = 10s, a = zero

Between t= 10s and t = 12s, a = -4 / 2 = -2 m/s²Over the whole 12s, average acceleration = ZERO

On a velocity-time graph, the area under the graph gives the displacement (the change in position).

Velocity (m/s)

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Displacement = 4 x 10 = 40 m

Note that although we can get velocity from a position-time graph, we cannot get position from a velocity-time graph. We can get displacement (change in position) but can’t get position. For example, imagine that we know a car has had a velocity of zero for 10 seconds. From that information we can know that it’s displacement is zero (its position has not changed) but we can’t actually know where the car is (its position).

Velocity (m/s)

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Δx = ½ x 4 x 10 = 20m

Velocity (m/s)

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First four seconds: Δx = 4 x 4 = 16mNext ten seconds: Δx = ½ x 4 x 10 = 20m

Total displacement = 16m + 20m = 36m

Velocity (m/s)

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An area below the time axis is a NEGATIVE AREA and corresponds to a NEGATIVE displacement i.e. a backwards displacement.

Δx = -4 x 8 = -32m

Velocity (m/s)

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First four seconds: Δx = ½ x -2 x 4 = -4mNext six seconds: Δx = -2 x 6 = -12m

Total displacement = -4m + -12m = -16m

Velocity (m/s)

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Total displacement = 8m + -8m = 0m

First four seconds: Δx = ½ x 4 x 4 = 8mNext four seconds: Δx = ½ x -4 x 4 = -8m

THINK about that result!

Velocity (m/s)

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Δx = 0.5 x 4 x 4 = 8m Δx = 0.5 x 4 x 2 = 4mΔx = 4 x 6= 24m

Total displacement = 8m + 24m + 4m = 36m

Equations of motion (1)Velocity

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Displacement = area on graph = ut

Total displacement Δx = ut + ½ (v-u)t = ut + ½vt - ½ut = ½(v+u)t

Displacement= area on graph = ½(v-u)t

v-u

= average velocity x time

Equations of motion (2)

You have already learnt that acceleration is the rate of change of velocity, or the change of velocity divided by the time taken

a = (v-u)/t

Equations of motion (3)

On the previous slide we hada = (v-u)/t

Rearranging… at = v – u

u + atv =

Equations of motion (4)

Three slides back we had displacement Δx = ut + ½ (v-u)t ……… (1)but we’ve also seen a=(v-u)/twhich means v-u = at

Substituting for v-u in equation (1) we get Δx = ut + ½ (at)t

Δx = ut + ½ at²

Equations of motion (all 5)In similar ways, we can find other equations of motion. Here are the five equations of motion you will find most useful, including the four we just found and one other which you may be able to find for yourself:

a = (v-u)/tv = u + at

Δx = ½ (v+u)t Δx = ut + ½ at² v² = u² + 2aΔx

Bonus equation, not often needed and not usually included in the ‘official’ list:

Δx = vt - ½ at²

Choosing the right equation of motion (1)

• Notice that in each equation of motion there is one quantity ‘missing’. For example, the equation v = u + at does not contain ‘Δx’. This is the clue that will help you choose which equation of motion may be appropriate to your problem.

• Identify the quantities given in the problem and the quantity that you are asked to find, then identify what quantity you are not given and not asked to find – that will tell you which equation to use.

Choosing the right equation of motion (2)• An example: A car that is traveling at 4 m/s experiences an

acceleration of 3m/s² for 5 s. What is its displacement during this time (assume the motion is linear).

• ‘Linear motion’ is straight line motion.• You are given u, a and t and asked to find Δx so the missing

quantity is v. The only equation of motion that does not mention v is…

Δx = ut + ½ at²•Substituting values into that equation:Δx = ut + ½ at²Δx = (4 m/s)· (5s) + ½ (3m/s²) ·(5s)² = 20m + ½ (3m/s²) ·(25s²) = 20m + ½ (75m) = 20m + 37.5m = 57.5m •We were asked to find the displacement which is a vector. Shouldn’t our answer indicate the direction as well as the magnitude? Our answer does include the direction as long as it is understood that the forward direction is positive.