mechyr1 chapter 9 :: constant acceleration · (i.e. gradient of velocity-time graph) the area under...

MechYr1 Chapter 9 :: Constant Acceleration

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RECAP :: Displacement-Time Graphs

Describe the motion of each object:

𝑠 (𝑚)

𝑡 (𝑠)

𝑠 (𝑚)

𝑡 (𝑠)

𝑠 (𝑚)

𝑡 (𝑠)

Object is stationary. Object is moving with constant velocity.

Object is accelerating.

Velocity is the rate of change of displacement (i.e. gradient of displacement-time graph)

Average Velocity =Displacement from starting point

Time taken

Average Speed =Total distance travelled

Time taken

The distinction is important. If you went out then some time later travelled back home, your average velocity is 0 because your eventual displacement is 0!

? ? ?

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Example (Exercise 9A Question 2)

𝑣 = 60𝑘𝑚ℎ−1. In the first 21

2hours, Khalid travels 60 × 2.5 = 150𝑘𝑚

One vertical square = 150 ÷ 6 = 25km

Total displacement = 71

2squares so 25 × 7

1

2= 187.5𝑘𝑚

a

b

?

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒=187.5𝑘𝑚

3.75ℎ= 50𝑘𝑚ℎ−1?

Exercise 9A

Pearson Stats/Mechanics Year 1Pages 132-133

RECAP :: Velocity-Time Graphs

Describe the motion of each object:

𝑣 (𝑚𝑠−1)

𝑡 (𝑠)

𝑣 (𝑚𝑠−1)

𝑡 (𝑠)

𝑣 (𝑚𝑠−1)

𝑡 (𝑠)

Object is stationary.Object is moving with constant velocity (as

change in velocity is 0).

Object has constant acceleration (as velocity is

increasing at constant rate).? ? ?

Acceleration is the rate of change of velocity (i.e. gradient of velocity-time graph)

The area under a velocity-time graph gives the distance travelled.?

Note: We’ll see later in Chapter 11 that when we differentiate displacement we get velocity, and therefore integrating velocity gives displacement. But we know that integrating finds the area under the graph.

?

Examples

The velocity-time graph shown is for a body which starts from rest, accelerates uniformly to a velocity of 8ms-1 in 2 seconds, maintains that velocity for a further 5 seconds then decelerates uniformly to rest. The entire journey takes 11 seconds. Find:a) The acceleration of the body during the initial part of the motionb) The deceleration of the body during the final part of the motionc) The total distance travelled by the body

𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =8

2= 4𝑚𝑠−2

𝐷𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = −8

4= −2𝑚𝑠−2

But deceleration (slowing down) is negative acceleration, so deceleration = 2ms-2

Distance travelled = area under graph

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =1

2× 5 + 11 × 8 = 64𝑚

a

b

?

? In case you’ve forgotten:

Area of trapezium= average of parallel sides× height between them

You’re welcome.

c

?

NB! If the graph includes negative velocities, you must calculate the areas under the graph separately to find the total distance (as negative velocities will give negative displacements)

Algebraic Example

[Textbook] A particle moves along a straight line. The particle accelerates uniformly from rest to a velocity of 8 ms-1 in 𝑇 seconds. The particle then travels at a constant velocity of 8 ms-1 for 5𝑇 seconds. The particle then decelerates uniformly to rest in a further 40 s.(a) Sketch a velocity-time graph to illustrate the motion of the particle.Give then the total displacement of the particle is 600m.(b) find the value of 𝑇.

𝑣 (𝑚𝑠−1)

𝑡 (𝑠)

8

𝑇 5𝑇 40

𝐴 𝐵

𝐶𝑂

Tip: Sometimes it’s easier to indicate the period of time that has passed (using arrows) rather than the time at the end of the interval.

Using area:

5𝑇 + 6𝑇 + 40

2× 8 = 600

44𝑇 + 160 = 600𝑇 = 10

a b

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Test Your Understanding

Edexcel M1 May 2013

You won’t likely have the knowledge for (d) yet…

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Exercise 9B


Edexcel M1 May 2012 Q4

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For (b), it may be helpful to know that:final velocity = initial velocity+ (time × acceleration)

Constant Acceleration Formulae (SUVAT equations)

When there is constant acceleration, there are a variety of formulae which relate the following 5 quantities:

𝒔: displacement𝒖: initial (starting) velocity𝒗: final velocity𝒂: acceleration𝒕: time

Velocity

Time

𝑣

𝑡

𝑢

It’s important you recognise these equations are for a specific interval of time. So the time 𝑡 is the duration of the period we’re considering, not necessarily the time since the object was moving.

These formulae are used to solve problems where the object is moving in a straight line with constant acceleration.

You are expected to be able to prove each “suvat” question using the above graph.

Using the gradient of the graph (which we know is acceleration):

𝑎 =𝑣−𝑢

𝑡→ ! 𝑣 = 𝑢 + 𝑎𝑡 (1)

Using the area under the graph (which we know gives distance):

! 𝑠 =𝑢+𝑣

2𝑡 (2)

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Memorisation Tip: This formula is effectively “distance = average speed × time” which you knew from GCSE.

SUVAT Equations

The other SUVAT equations can be derived from the first two:

Eliminating 𝑡 – sub from (1) into (2):

𝑡 =𝑣 − 𝑢

𝑎

𝑠 =𝑢 + 𝑣

2

𝑣 − 𝑢

𝑎

𝑠 =𝑣2 − 𝑢2

2𝑎

! 𝑣2 = 𝑢2 + 2𝑎𝑠 (4)

Eliminating 𝑢 – sub from (1) into (2):𝑢 = 𝑣 − 𝑎𝑡

𝑠 =𝑣 − 𝑎𝑡 + 𝑣

2𝑡

=2𝑣 − 𝑎𝑡

2𝑡

! 𝑠 = 𝑣𝑡 −1

2𝑎𝑡2

Fro Note: I have never seen an exam question that uses this 𝑠𝑢𝑣𝑎𝑡 formula.

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Eliminating 𝑣 - sub from (1) into (2):

𝑠 =𝑢 + 𝑢 + 𝑎𝑡

2𝑡

=2𝑢 + 𝑎𝑡

2𝑡

! 𝑠 = 𝑢𝑡 +1

2𝑎𝑡2 (3)

?

Fro Note: Because this is quadratic in 𝑡, we typically end up with two different possible times.

Examples

A stone slides in a straight line across a horizontal sheet of ice. It passes a point, A with velocity 14ms-1 and a point, B 2.5 seconds later. Assuming the deceleration is uniform and that AB = 30m, find:a) The decelerationb) The velocity at Bc) How long after passing A the stone comes to rest

a b

Always write each letter of suvat, putting a “?” for any quantities you need to find.

𝑢 = 14𝑚𝑠−1 𝑣1 = ? 𝑣2 = 0

A B C

𝑠1 = 30𝑚𝑡1 = 2.5𝑠

𝑡2 = ?

Diagram:

𝑢 = 14𝑚𝑠−1

𝑠 = 30𝑚𝑡1 = 2.5𝑠𝑎 = ?

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

30 = 14 2.5 +1

2𝑎 2.5 2

𝑎 = −1.6𝑚𝑠−2

The deceleration is 1.6ms-2.

?

We can choose which equation to use for 𝑣1.

𝑢 = 14𝑚𝑠−1

𝑡 = 2.5𝑠𝑠 = 30𝑚𝑣1 = ?

𝑠 =(𝑢 + 𝑣)

2𝑡

30 =14 + 𝑣1

22.5

𝑣1 = 10𝑚𝑠−1

Safer to use given values

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c We can use displacement AC or BC.AC:𝑢 = 14𝑚𝑠−1

𝑣2 = 0 𝑟𝑒𝑠𝑡𝑡2 = ?𝑎 = −1.6𝑚𝑠−2

𝑣 = 𝑢 + 𝑎𝑡

0 = 14 − 1.6𝑡2

𝑡2 = 8.75𝑠

?

𝑠 = 32𝑚𝑡 = 4𝑠 𝑎𝑛𝑑 𝑡 = 8𝑠𝑎 = −2𝑚𝑠−2

𝑢 = 12𝑚𝑠−1

𝑣 = ?

We can choose which equation to use.

𝑣 = 𝑢 + 𝑎𝑡At 𝑡 = 4𝑠:𝑣1 = 12 − 2 4

𝑣1 = 4𝑚𝑠−1 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵

At 𝑡 = 8𝑠:𝑣2 = 12 − 2 8

𝑣2 = −4𝑚𝑠−1 𝑜𝑟 𝑣2 = 4𝑚𝑠−1𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝐵𝐴

Be careful with positive and negative directions

Diagram:

𝑢 = 12𝑚𝑠−1

𝑡 = ?𝑠 = 32𝑚𝑎 = −2𝑚𝑠−2

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

32 = 12𝑡 −2

2𝑡2

Rearrange and solve the quadratic:𝑡2 − 12𝑡 + 32 = 0𝑡 − 8 𝑡 − 4 = 0⇒ 𝑡 = 4𝑠 𝑎𝑛𝑑 𝑡 = 8𝑠

Examples – Deceleration Leading to a Change of Direction

A particle travels with uniform deceleration 2ms-2 in a horizontal line. The points A and B lie on the line and AB = 32m. At time t = 0, the particle passes through A with velocity 12ms-1 in the direction AB. Find:a) The values of t when the particle is at Bb) The velocity of the particle for each of these values of t.

a b

?

𝑢 = 12𝑚𝑠−1𝑡1 = ? , 𝑣1 = ?𝑡2 = ? , 𝑣2 = ?

A B

𝐴𝐵 = 32𝑚

𝑎 = 2𝑚𝑠−2

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Edexcel M1 May 2013 Q4

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Exercise 9C and Exercise 9D

Pearson Stats/Mechanics Year 1Pages 140-141 and pages 145 - 146

Vertical Motion Under Gravity

Famously, when the Apollo 15 landed on the moon, astronaut David Scott conducted a famous demonstration in which a hammer and feather were released at the same time. As anticipated, they hit the ground at the same time!

If there is no air resistance, then the acceleration of objects under gravity, regardless of mass, is constant.

The downwards acceleration under gravity is 𝑔 = 9.8 ms-2.

Important Note: It’s important you use 9.8 and not 10 or 9.81, which is often used in other exam boards/Physics. Also note that given we’re using the value of 𝑔 to 2 significant figures, any subject value calculated should also be given to 2 significant figures.

Example

A ball is thrown vertically upwards with a velocity of 14.7ms-1 from a platform 19.6m above the ground. Find:a) The time taken for the ball to reach the groundb) The velocity of the ball when it hits the ground

𝑠 = −19.6𝑚𝑢 = 14.7𝑚𝑠−1

𝑎 = −9.8𝑚𝑠−2,𝑡 = ?

At this stage, it’s hugely important you consider what direction is considered as ‘positive’, and mark it next to your suvat values. If ‘up’ was positive, then 𝑎 = −9.8. If ‘down’ is positive, then 𝑎 = +9.8. The direction does not matter provided that you are consistent with each letter of suvat, but convention is that we make as many values positive as possible.

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

−19.6 = 14.7𝑡 +1

2(−9.8)(𝑡2)

4.9𝑡2 − 14.7𝑡 − 19.6 = 0𝑡 = 4𝑠 𝑜𝑟 𝑡 = −1𝑠

a

b The ball reaches the ground at 𝑡 = 4𝑠.𝑣 =? 𝑣2 = 𝑢2 + 2𝑎𝑠𝑠 = −19.6 𝑣2 = 14.72 + 2(−9.8)(−19.6)

𝑎 = −9.8 𝑣2 =2401

4

𝑢 = 14.7𝑣 = 25𝑚𝑠−1 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠 (2 sf)

As per previous slide, quote only to 2 significant figures. You may be penalised if you quote more!

Invalid as 𝑡 > 0

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𝑢 = 14.7𝑚𝑠−1

𝑔 = 9.8𝑚𝑠−2

s = 19.6𝑚

A common type of question…

A ball is projected vertically upwards from ground level at a speed of 20 ms-1.Determine the amount of time the ball is at least 10m above ground level.

10m

?a 𝑠 = 10, 𝑢 = 20, 𝑎 = −9.8, 𝑡 =?? ? ?

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

10 = 20𝑡 − 4.9𝑡2

𝑡 = 0.5834 𝑜𝑟 𝑡 = 3.4983Therefore time above 10m:3.4983 − 0.5834 = 2.9 s (2sf)

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Calculator Tip: Be sure to use the quadratic solver on your calculator (within ‘Equation’ mode on the ClassWiz).

𝑔

When Two Particles are in Motion

Two stones are thrown from the same point at the same time - one vertically upwards with speed 30ms-1 and one vertically downwards at 30ms-1. Find how far apart the stones are after 3 seconds.

? 𝑠 = ?, 𝑢 = 30, 𝑎 = −9.8, 𝑡 = 3 ?? ?

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

𝑠 = 30 3 − 4.9 32

𝑠 = 45.9𝑚

?A

B

𝑢𝐵 = 30𝑚𝑠−1

𝑢𝐴 = 30𝑚𝑠−1

𝑔 = 9.8𝑚𝑠−2

A:

? 𝑠 = ?, 𝑢 = 30, 𝑎 = 9.8, 𝑡 = 3?? ?B:

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

𝑠 = 30 3 + 4.9 32

𝑠 = 134.1𝑚

Total distance = 45.9 + 134.1 = 180m

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Edexcel M1 May 2013(R) Q4

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Exercise 9E


mechyr1 chapter 9 :: constant acceleration · (i.e. gradient of velocity-time graph) the area under...

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