Vectors and Equations of Motion (suvat) P5b(ii) You will find out about: How to use equations of motion

Vectors and Equations of Motion (suvat) P5b(ii)

You will find out about:How to use equations of motion

www.PhysicsGCSE.co.uk

http://www.physicsgcse.co.uk/

Average SpeedSpeed usually always changes during a journey. If you think about a car journey there are many obstacles that allow cause the driver to change the speed. This could be slowing down for traffic lights or a pedestrian crossing or speeding up if joining a Motorway.

It is practical then to calculate average speed for a journey:average speed = with unit of m/s

The Shanghai Maglev Train runs from Longyang Road station in Pudong to Pudong International Airport in China. At full speed the journey takes 7 minutes and 20 seconds to complete the distance of 30 km.

Calculate its average speed:

average speed =

Convert 7 minutes 20 seconds to seconds = (7x60)+20 = 500 sConvert 30 km into metres = 30,000 m

average speed = = = 60 m/s

This is approximately 216km/hTop speeds can reach in excess of 431 km/h but of course the Maglev train does not run at that speed during the entire journey for example when it starts its journey it needs to build up to that speed.



Equations of MotionWhen an object accelerates in a straight line the equations of motion are used to find out more about how it moves.You need to remember ‘suvat’:s = distance travelled , mu = initial velocity, m/sv = final velocity, m/sa = acceleration, m/s2

t = time, s


REMEMBER:It is very easy to confuse u and v. v and f (for final…) start with the same sound


Equation 1

You learnt in P2 that:Average speed= and by using the speed triangle from P3we know that distance = average speed x time.

Distance becomes s Average speed is And time is just t.Put it together and we get:distance = average speed x time

s = x t

www.PhysicsGCSE.co.uk s = distance travelled , mu = initial velocity, m/sv = final velocity, m/sa = acceleration, m/s2

t = time, s

Example Question:A train accelerates uniformly from 40m/s to 70m/s in 4 seconds. How far does it travel?From the question write out ‘suvat’ and put in the numbers:s = need to find out (remember distance is now s)u = 40m/sv = 70m/sa = do not knowt = 4 secondsIt does not matter that we do not know a as Equation 1 does not need a:s =

s = = 220m


Equation 2

The next equation you need to know is:Final speed = initial speed + (acceleration x time)orv = u + at


t = time, s

Example Question:An aeroplane moving at 60m/s starts to accelerate uniformly at 12m/s2. How fast is it moving after 20 seconds?Again use ‘suvat’ and input the numbers:s = do not knowu = 60m/sv = need to work outa = 12m/s2

t = 20sIt does not matter that we do not know s as Equation 2 does not use it:v = u + atv = 60 + (12x20)v = 300m/s


A little more about Equations 1 and 2You also need to be able to re-arrange Equations 1 and 2 and know how they were derived.

Equation 1You previously learned that:Average speed = So using the speed triangle we know that:

average speed = or =

By multiplying both sides by t we then get:s = x t


Equation 2You previously learned that:acceleration = ora = By multiplying both sides by t:at = v-uand then adding u to both sides we get:v = u + at

This is Equation 1

This is Equation 2

Practice re-arranging these equations so you can show how to derive them in the form they are shown in the exam paper.


Difficult Questions using Equations 1 and 2

Difficult Question 1:A bullet reached a speed of 400m/s after accelerating uniformly at 20,000m/s2 for 0.001 second. How fast was it going before accelerating?Using ‘suvat’:s = do not knowu = need to find outv = 400 m/sa = 20,000 m/s2

t = 0.001 sWe cannot use equation 1 as we do not know s. We can however, use equation 2:v = u + atThis is more difficult because we need to work out u rather than v.Most students find it easier to re-arrange the equation at the end:v = u + atplug numbers in:400 = u + (20,000 x 0.001)400 = u + 20

now we can re-arrange :

400 – 20 = u380 m/s = u


t = time, s

If your math is good you could re-arrange the equation from the start so that:u = v - atBUT if you make a mistake here you will get the wrong answer!

Difficult Question 2:A toy dog reaches a maximum speed of 0.08m/s. If it starts from rest and travels a distance of 0.30m how long does it take?Using ‘suvat’:s = 0.30 mu = 0 m/s (remember it is starting from rest)v = 0.08 m/sa = do not knowt = need to find outThis time we will use equation 1 as we will be using s, u, v and t:s = x tPlug numbers in:0.30 = x t0.30 = 0.04 x t

Divide both sides by 0.04:

7.5 seconds = t

If your math is good you could re-arrange the equation from the start so that:t = BUT if you make a mistake here you will get the wrong answer!


More equations of motion for uniform acceleration!

Equation 1 shows us that: s = x t


and we know that Equation 2 is: v = u + at

For Equation 1 we can substitute u + at for v because they are equal:s = x tnow change v to u + ats = and if we tidy the equation we get to:s = ut + ½ at2

You can see that v is now swapped for u+at

THIS IS EQUATION 3!

Equation 1 shows us that: s = x t

and we know that Equation 2 is: v = u + at

We car re-arrange Equation 2 so that:t =

Again we can substitute Equation 2 into Equation 1:s = x tnow change t to s = x and if we tidy the equation we get to:s = or v2 = u2 + 2 as

You can see that t is now swapped for

THIS IS EQUATION 4!


Re-cap of Equations

Equation 1: s = x t

Equation 2: v = u + at

Equation 3: s = ut + ½ at2

Equation 4: v2 = u2 + 2 as


t = time, s

If we know three of the ‘suvat’ quantities then we can find out the other two using the four Equations.


Taking stock…

Equation 1: s = x t





t = time, sQuestion:Usain Bolt reaches a speed of 12.27 m/s from rest after running 45 m. Find his acceleration.

Using ‘suvat’:

s = 45 mu = 0 m/sv = 12.27 m/sa = need to work outt = do not know

Let us look at the four Equations to determine which we can use:

no, because we do not know t

no, because we do not know t and does not use s

no, because we do not know t

yes!!

Using Equation 4:v2 = u2 + 2 as12.272 = 02 + (2 x a x 45)150.5529 = 90 x a

= a

1.67281 m/s2 = a

Which Equation shall we use?


Does Gravity play a part?In a word, yes!If we ignore air resistance all objects accelerate towards the Earth at 10m/s2 due to gravity.

But an object going upwards decelerates at 10m/s2. As it is going in the opposite direction to the Earth its acceleration is -10m/s2 and at its highest point its velocity, v, would be 0m/s.


This trampolinist jumping upward is moving in the opposite direction of the Earth. Therefore their acceleration is -10m/s2.

When they reach their highest point v will be 0m/s.

On their way back down, and moving in the direction towards Earth, their acceleration will be 10m/s2.

Question:This trampolinist jumped a second time vertically upward at 12.5m/s. How high did they go?Using ‘suvat’:s= need to find outu = 12.5 m/sv = 0 m/sa = -10 m/s2

t = do not know

Which Equation shall we use?

Equation 1: s = x t




No because we do not know t

Equation 4:v2 = u2 + 2 as02 = 12.52 + (2x-10 x s)

take away 12.52 on both sides

02 – 12.52 = (2 x -10 x s)-156.25 = -20 x s

Divide -20 both sides

= s

7.8125 m = s

No, because we want to find out s and we also do not know t

No because we do not know t

yes!!


Questions

1. Give two reasons why a speed of a lorry may change during its trip to a warehouse.

2. A cyclist started cycling a bike with an acceleration of 3.5m/s2. How fast was it going after 20 seconds? Is this possible?

3. A Ferrari accelerated at 8m/s2 for 6 seconds reaching a top speed of 345 km/h. How fast was it moving before accelerating in m/s? How far did it travel?

4. A cannonball was dropped from a chalk pit that was 120 metres high. With what speed did it hit the ground below?



Questions1. Give two reasons why a speed of a lorry may change during its trip to a

warehouse. Traffic lights, roundabouts and congestion could slow it down. Clear roads and faster roads could enable the lorry to be driven faster.

2. A cyclist started cycling a bike with an acceleration of 3.5m/s2. How fast was it going after 20 seconds? Is this possible? v=u+at v=0+3.5x20 v=70m/s. Not possible for the average human being!

3. A Ferrari accelerated at 8m/s2 for 6 seconds reaching a top speed of 345 km/h. How fast was it moving before accelerating in m/s? How far did it travel? Use v=u+at so 345=?+(8x6) u=297m/s then use s=x t = x6 = 1926m (NOTE: could have used the equation s=ut+ ½ at2 and you would have got the same answer!)

4. A cannonball was dropped from a chalk pit that was 120 metres high. With what speed did it hit the ground below? use v2=u2+2as = v2=0+(2x10x120) then v = √2400 = 48.99 m/s



Documents

Vectors and Equations of Motion (suvat) P5b(ii) You will find out about: How to use equations of motion