Kinematics of a particle moving in a straight line

KINEMATICS OF A PARTICLE MOVING IN

A STRAIGHT LINE

MECHANICS 1 6677 (M1)

JAN 27, 14

21m

s-1

Introduction• This chapter you will learn the SUVAT

equations

• These are the foundations of many of the Mechanics topics

• You will see how to use them to use many types of problem involving motion

TEACHINGS FOR EXERCISE 2A

Kinematics of a Particle moving in a Straight Line

You will begin by learning two of the SUVAT equations

s = Displacement (distance)u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time

2A

𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛=h𝑐 𝑎𝑛𝑔𝑒 𝑖𝑛𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦h𝑐 𝑎𝑛𝑔𝑒 𝑖𝑛𝑡𝑖𝑚𝑒

𝑎=𝑣−𝑢𝑡

𝑎𝑡=𝑣−𝑢

𝑎𝑡+𝑢=𝑣𝑣=𝑢+𝑎𝑡

Replace with the appropriate letters. Change in velocity

= final velocity – initial velocity

Multiply by t

Add u

This is the usual form!

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑚𝑜𝑣𝑒𝑑=𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑠𝑝𝑒𝑒𝑑×𝑡𝑖𝑚𝑒

𝑠=(𝑢+𝑣2 )𝑡

Replace with the

appropriate letters




2A

𝑣=𝑢+𝑎𝑡𝑠=(𝑢+𝑣

2 )𝑡

You need to consider using negative numbers in some cases

P Q

Positive direction

O4m 3m

2.5ms-1 6ms-1

If we are measuring displacements from O, and left to right is the positive direction…

For particle P: For particle Q:𝑠=−4𝑚

𝑣=2.5𝑚𝑠−1𝑠=3𝑚

𝑣=−6𝑚𝑠−1The particle is to the left of

the point O, which is the negative direction

The particle is moving at 2.5ms-1 in the positive

direction

The particle is to the right of the

point O, which is the positive

direction

The particle is moving at 6ms-1 in the negative

direction




2A


2 )𝑡

A particle is moving in a straight line from A to B with constant acceleration 3ms-2. Its speed at A is 2ms-1 and it takes 8 seconds to move from A to B. Find:a) The speed of the particle at Bb) The distance from A to B

𝑠=? 𝑢=2 𝑣=? 𝑎=3 𝑡=8A B

2ms-1

Start with a diagram

Write out ‘suvat’ and fill in what you

knowFor part a) we need to calculate v, and we know u, a and

t…

𝑣=𝑢+𝑎𝑡𝑣=2+(3×8)

𝑣=26𝑚𝑠−1

Fill in the values you

knowRemember to include

units!

You always need to set up the question in this way. It makes it much easier to figure

out what equation you need to use (there will be more to learn than just these two!)




2A


2 )𝑡

A particle is moving in a straight line from A to B with constant acceleration 3ms-2. Its speed at A is 2ms-1 and it takes 8 seconds to move from A to B. Find:a) The speed of the particle at B – 26ms-1

b) The distance from A to B

𝑠=? 𝑢=2 𝑣=? 𝑎=3 𝑡=8A B

2ms-1

𝑣=26For part b) we need to calculate s, and we know u, v and

t…


𝑠=( 2+262 )×8𝑠=(14 )×8

𝑠=112𝑚

Fill in the values you

knowShow

calculations

Remember the units!




2A


2 )𝑡

A cyclist is travelling along a straight road. She accelerates at a constant rate from a speed of 4ms-1 to a speed of 7.5ms-1 in 40 seconds. Find:a) The distance travelled over this 40 secondsb) The acceleration over the 40 seconds

4ms-1 7.5ms-1

𝑠=? 𝑢=4 𝑎=? 𝑡=40𝑣=7.5

Draw a diagram (model the cyclist

as a particle)Write out ‘suvat’

and fill in what you know

𝑠=(𝑢+𝑣2 )𝑡 We are calculating

s, and we already know u, v and t…

𝑠=( 4+7.52 )×40𝑠=230𝑚

Sub in the values you

know

Remember units!




2A


2 )𝑡

A cyclist is travelling along a straight road. She accelerates at a constant rate from a speed of 4ms-1 to a speed of 7.5ms-1 in 40 seconds. Find:a) The distance travelled over this 40 seconds – 230mb) The acceleration over the 40 seconds

4ms-1 7.5ms-1

𝑠=230 𝑢=4 𝑎=? 𝑡=40𝑣=7.5

Draw a diagram (model the cyclist

as a particle)Write out ‘suvat’

and fill in what you know

For part b, we are calculating a, and

we already know u, v and t…


knowSubtract 4

𝑣=𝑢+𝑎𝑡7.5=4+40𝑎7.5=40𝑎

Divide by 40𝑎=0.0875𝑚𝑠−2

𝑠=?




2A


2 )𝑡

A particle moves in a straight line from a point A to B with constant deceleration of 1.5ms-2. The speed of the particle at A is 8ms-1 and the speed of the particle at B is 2ms-1. Find:a) The time taken for the particle to get from A to Bb) The distance from A to B

8ms-1 2ms-1

𝑠=?𝑢=8 𝑎=−1.5𝑡=?𝑣=2

Draw a diagram


knowAs the particle is

decelerating, ‘a’ is negative𝑣=𝑢+𝑎𝑡

2=8−1.5 𝑡−6=−1.5 𝑡4=𝑡


knowSubtract 8

Divide by -1.5

A B




2A


2 )𝑡

A particle moves in a straight line from a point A to B with constant deceleration of 1.5ms-2. The speed of the particle at A is 8ms-1 and the speed of the particle at B is 2ms-1. Find:a) The time taken for the particle to get from A to B – 4 secondsb) The distance from A to B

8ms-1 2ms-1

𝑠=?𝑢=8 𝑎=−1.5𝑡=?𝑣=2

Draw a diagram


knowAs the particle is

decelerating, ‘a’ is negative


know

𝑡=4


𝑠=( 8+22 )×4𝑠=20𝑚

Calculate the answer!

A B




2A


2 )𝑡

After reaching B the particle continues to move along the straight line with the same deceleration. The particle is at point C, 6 seconds after passing through A. Find:a) The velocity of the particle at Cb) The distance from A to C

8ms-1 2ms-1

𝑠=?𝑢=8 𝑎=−1.5𝑣=? 𝑡=6A B C

?Update the

diagram

Write out ‘suvat’ using

points A and C

𝑣=𝑢+𝑎𝑡𝑣=8−(1.5×6)

𝑣=−1𝑚𝑠−1

Sub in the valuesWork it

out!

As the velocity is negative, this means the particle has now changed direction

and is heading back towards A! (velocity has a direction as well as a

magnitude!)The velocity is 1ms-1 in the direction C to A…




2A


2 )𝑡

After reaching B the particle continues to move along the straight line with the same deceleration. The particle is at point C, 6 seconds after passing through A. Find:a) The velocity of the particle at C - -1ms-1

b) The distance from A to C8ms-1 2ms-1

𝑠=?𝑢=8 𝑎=−1.5𝑣=? 𝑡=6A B C

?Update the

diagram

Write out ‘suvat’ using

points A and C𝑣=−1


𝑠=( 8−12 )×6𝑠=21𝑚

Sub in the values

Work it out!

It is important to note that 21m is the distance from A to C only…

The particle was further away before it changed direction, and has in total travelled further than 21m…




2A


2 )𝑡

A car moves from traffic lights along a straight road with constant acceleration. The car starts from rest at the traffic lights and 30 seconds later passes a speed trap where it is travelling at 45 kmh-1. Find:a) The acceleration of the carb) The distance between the traffic lights and the speed-trap.0ms-1 45kmh-1

Lights Trap

Standard units to use are metres and seconds, or kilometres and hours In this case, the time is in seconds and the speed is in

kilometres per hour We need to change the speed into metres per second first!

Draw a diagram

45𝑘𝑚h−1

45,000𝑚h−1

12.5𝑚𝑠−1

Multiply by 1000 (km to m)

Divide by 3600 (hours to seconds)




2A


2 )𝑡


Lights TrapDraw a diagram

= 12.5ms-1


know𝑠=?𝑢=0 𝑎=? 𝑡=30𝑣=12.5

𝑣=𝑢+𝑎𝑡12.5=0+30𝑎512𝑚𝑠−2=𝑎

Sub in the values

Divide by 30

You can use exact

answers!




2A


2 )𝑡


Lights TrapDraw a diagram

= 12.5ms-1


know𝑠=?𝑢=0 𝑎=? 𝑡=30𝑣=12.5𝑎=

512


𝑠=( 0+12.52 )×30𝑠=187.5𝑚

Sub in values

Work it out!

TEACHINGS FOR EXERCISE 2B


You can also use 3 more formulae linking different combination of

‘SUVAT’, for a particle moving in a straight line with constant

acceleration

2B


2 )𝑡

𝑣=𝑢+𝑎𝑡𝑣−𝑢=𝑎𝑡𝑣−𝑢𝑎 =𝑡


𝑠=(𝑢+𝑣2 )

𝑠=(𝑣2−𝑢2

2𝑎 )2𝑎𝑠=𝑣2−𝑢2

𝑢2+2𝑎𝑠=𝑣2

𝑣2=𝑢2+2𝑎𝑠

𝑣2=𝑢2+2𝑎𝑠

Subtract u

Divide by a

Replace t with the expression above

Multiply numerators and denominators

Multiply by 2a

Add u2

This is the way it is usually written!

(𝑣−𝑢𝑎 )




acceleration

2B


2 )𝑡𝑣2=𝑢2+2𝑎𝑠


𝑠=(𝑢+𝑢+𝑎𝑡2 )𝑡

𝑠=( 2𝑢+𝑎𝑡2 )𝑡

𝑠=(𝑢+ 12𝑎𝑡 )𝑡

𝑠=𝑢𝑡+ 12 𝑎𝑡2

Replace ‘v’ with ‘u + at’

Group terms on the numerator

Divide the numerator by 2

Multiply out the bracket





acceleration

2B


2 )𝑡𝑣2=𝑢2+2𝑎𝑠


𝑣=𝑢+𝑎𝑡𝑣−𝑎𝑡=𝑢


𝑠=(𝑣−𝑎𝑡)𝑡+ 12 𝑎𝑡2

𝑠=𝑣𝑡−𝑎𝑡2+ 12 𝑎𝑡2

𝑠=𝑣𝑡− 12 𝑎𝑡2


Subtract ‘at’

Replace ‘u’ with ‘v - at’ from above’

Multiply out the bracket

Group up the at2 terms


You can also use 3 more formulae linking different

combination of ‘SUVAT’, for a particle moving in a straight

line with constant acceleration

2B

𝑣=𝑢+𝑎𝑡 𝑠=(𝑢+𝑣2 )𝑡

𝑣2=𝑢2+2𝑎𝑠𝑠=𝑢𝑡+ 12 𝑎𝑡2


A particle is moving in a straight line from A to B with constant acceleration 5ms-2. The velocity of the particle at A is 3ms-1 in the direction AB. The velocity at B is 18ms-1 in the same direction. Find the distance from A to B.

3ms-1 18ms-1

A BDraw a diagram

𝑠=?𝑢=3 𝑎=5 𝑡=?𝑣=18Write out ‘suvat’

with the information

given

𝑣2=𝑢2+2𝑎𝑠182=32+2(5)𝑠

324=9+10 𝑠315=10 𝑠

31.5𝑚=𝑠

Replace v, u and a

Work out terms

Subtract 9

Divide by 10

We are calculating s,

using v, u and a





2B




A car is travelling along a straight horizontal road with a constant acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it passes a pillar box. 12 seconds later the car passes a lamp post. Find:a) The distance between the pillar box and the lamp postb) The speed with which the car passes the lamp post

8ms-1

Pillar Box

Lamp Post

𝑠=?𝑢=8 𝑎=0.75𝑡=12𝑣=?

Draw a diagram

Write out ‘suvat’ with the

information given


using u, a and t𝑠=𝑢𝑡+ 12 𝑎𝑡2

𝑠=(8×12)+ 12 (0.75×122)

𝑠=150𝑚

Replace u, a and t

Calculate





2B




A car is travelling along a straight horizontal road with a constant acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it passes a pillar box. 12 seconds later the car passes a lamp post. Find:a) The distance between the pillar box and the lamp post – 150mb) The speed with which the car passes the lamp post

8ms-1

Pillar Box

Lamp Post

𝑠=?𝑢=8 𝑎=0.75𝑡=12𝑣=?

Draw a diagram


information given

We are calculating v,

using u, a and tReplace u, a

and t

Calculate

𝑣=𝑢+𝑎𝑡𝑣=8+(0.75×12)

𝑣=17𝑚𝑠−1

Often you can use an answer you have calculated later on in the same question. However, you must

take care to use exact values and not rounded answers!





2B




A particle is moving in a straight horizontal line with constant deceleration 4ms-2. At time t = 0 the particle passes through a point O with speed 13ms-1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through Ab) The total time the particle is beyond Ac) The time taken for the particle to return to O

13ms-1

O A𝑠=20𝑢=13 𝑎=−4𝑡=?𝑣=?

Draw a diagram


information given

We are calculating t,

using s, u and a𝑠=𝑢𝑡+ 12 𝑎𝑡2

20=(13)𝑡+ 12 (−4 )𝑡2

20=13 𝑡−2𝑡 2

2 𝑡2−13 𝑡+20=0(2 𝑡−5)(𝑡−4 )=0

𝑡=2.5𝑜𝑟 4

Replace s, u and a

Simplify terms

Rearrange and set equal to 0

Factorise (or use the quadratic formula…)

We have 2 answers. As the acceleration is negative, the particle passes through A, then changes direction and

passes through it again!





2B




A particle is moving in a straight horizontal line with constant deceleration 4ms-2. At time t = 0 the particle passes through a point O with speed 13ms-1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through A – 2.5 and 4 secondsb) The total time the particle is beyond Ac) The time taken for the particle to return to O

13ms-1

O A𝑠=20𝑢=13 𝑎=−4𝑡=?𝑣=?

Draw a diagram


information given

We are calculating t,

using s, u and aThe particle passes through A at 2.5

seconds and 4 seconds, so it was beyond A for 1.5 seconds…





2B




A particle is moving in a straight horizontal line with constant deceleration 4ms-2. At time t = 0 the particle passes through a point O with speed 13ms-1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through A – 2.5 and 4 secondsb) The total time the particle is beyond A – 1.5 secondsc) The time taken for the particle to return to O

13ms-1

O A𝑠=20𝑢=13 𝑎=−4𝑡=?𝑣=?

Draw a diagram


information given

The particle returns to O when s = 0

𝑠=0


0=(13)𝑡+(−2)𝑡 2

0=13 𝑡−2𝑡 2

2 𝑡2−13 𝑡=0𝑡 (2𝑡−13)=0

Replace s, u and a

𝑡=0𝑜𝑟 6.5

Simplify

Rearrange

FactoriseThe particle is at O when t =

0 seconds (to begin with) and is at O again when t =

6.5 seconds





2B




A particle is travelling along the x-axis with constant deceleration 2.5ms-2. At time t = O, the particle passes through the origin, moving in the positive direction with speed 15ms-1. Calculate the distance travelled by the particle by the time it returns to the origin.15ms-1

O X

Draw a diagram

The total distance travelled will be double the distance the particle reaches from O (point X)

At X, the velocity is 0𝑠=? 𝑢=15 𝑎=−2.5𝑡=?𝑣=0

𝑣2=𝑢2+2𝑎𝑠02=152+2(−2.5) 𝑠0=225−5 𝑠5 𝑠=225𝑠=45𝑚

Replace v, u and a

Simplify

Add 5s

Divide by 5

𝑇𝑜𝑡𝑎𝑙 𝑠=90𝑚45m is the distance from O to X. Double it for the total distance travelled


using u, v and a

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Kinematics of a particle moving in a straight line