Kinematics of a particle moving in a straight line
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Transcript of 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
Introductionβ’ This chapter you will learn the SUVAT
equations
β’ These are the foundations of many of the Mechanics topics
β’ 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
π΄ππππππππ‘πππ=hπ ππππ πππ£ππππππ‘π¦hπ ππππ πππ‘πππ
π=π£βπ’π‘
ππ‘=π£βπ’
ππ‘+π’=π£π£=π’+ππ‘
Replace with the appropriate letters. Change in velocity
= final velocity β initial velocity
Multiply by t
Add u
This is the usual form!
π·ππ π‘πππππππ£ππ=π΄π£ππππππ ππππΓπ‘πππ
π =(π’+π£2 )π‘
Replace with the
appropriate letters
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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 directionβ¦
For particle P: For particle Q:π =β4π
π£=2.5ππ β1π =3π
π£=β6ππ β1The 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
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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
π =? π’=2 π£=? π=3 π‘=8A B
2ms-1
Start with a diagram
Write out βsuvatβ and fill in what you
knowFor part a) we need to calculate v, and we know u, a and
tβ¦
π£=π’+ππ‘π£=2+(3Γ8)
π£=26ππ β1
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!)
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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 β 26ms-1
b) The distance from A to B
π =? π’=2 π£=? π=3 π‘=8A B
2ms-1
π£=26For part b) we need to calculate s, and we know u, v and
tβ¦
π =(π’+π£2 )π‘
π =( 2+262 )Γ8π =(14 )Γ8
π =112π
Fill in the values you
knowShow
calculations
Remember the units!
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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
π =? π’=4 π=? π‘=40π£=7.5
Draw a diagram (model the cyclist
as a particle)Write out βsuvatβ
and fill in what you know
π =(π’+π£2 )π‘ We are calculating
s, and we already know u, v and tβ¦
π =( 4+7.52 )Γ40π =230π
Sub in the values you
know
Remember units!
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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 β 230mb) The acceleration over the 40 seconds
4ms-1 7.5ms-1
π =230 π’=4 π=? π‘=40π£=7.5
Draw a diagram (model the cyclist
as a particle)Write out βsuvatβ
and fill in what you know
For part b, we are calculating a, and
we already know u, v and tβ¦
Sub in the values you
knowSubtract 4
π£=π’+ππ‘7.5=4+40π7.5=40π
Divide by 40π=0.0875ππ β2
π =?
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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
π =?π’=8 π=β1.5π‘=?π£=2
Draw a diagram
Write out βsuvatβ and fill in what you
knowAs the particle is
decelerating, βaβ is negativeπ£=π’+ππ‘
2=8β1.5 π‘β6=β1.5 π‘4=π‘
Sub in the values you
knowSubtract 8
Divide by -1.5
A B
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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 β 4 secondsb) The distance from A to B
8ms-1 2ms-1
π =?π’=8 π=β1.5π‘=?π£=2
Draw a diagram
Write out βsuvatβ and fill in what you
knowAs the particle is
decelerating, βaβ is negative
Sub in the values you
know
π‘=4
π =(π’+π£2 )π‘
π =( 8+22 )Γ4π =20π
Calculate the answer!
A B
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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
π =?π’=8 π=β1.5π£=? π‘=6A B C
?Update the
diagram
Write out βsuvatβ using
points A and C
π£=π’+ππ‘π£=8β(1.5Γ6)
π£=β1ππ β1
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 Aβ¦
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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
π =?π’=8 π=β1.5π£=? π‘=6A B C
?Update the
diagram
Write out βsuvatβ using
points A and Cπ£=β1
π =(π’+π£2 )π‘
π =( 8β12 )Γ6π =21π
Sub in the values
Work it out!
It is important to note that 21m is the distance from A to C onlyβ¦
The particle was further away before it changed direction, and has in total travelled further than 21mβ¦
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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
45ππhβ1
45,000πhβ1
12.5ππ β1
Multiply by 1000 (km to m)
Divide by 3600 (hours to seconds)
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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 TrapDraw a diagram
= 12.5ms-1
Write out βsuvatβ and fill in what you
knowπ =?π’=0 π=? π‘=30π£=12.5
π£=π’+ππ‘12.5=0+30π512ππ β2=π
Sub in the values
Divide by 30
You can use exact
answers!
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
π£=π’+ππ‘π =(π’+π£
2 )π‘
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 TrapDraw a diagram
= 12.5ms-1
Write out βsuvatβ and fill in what you
knowπ =?π’=0 π=? π‘=30π£=12.5π=
512
π =(π’+π£2 )π‘
π =( 0+12.52 )Γ30π =187.5π
Sub in values
Work it out!
TEACHINGS FOR EXERCISE 2B
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different combination of
βSUVATβ, for a particle moving in a straight line with constant
acceleration
2B
π£=π’+ππ‘π =(π’+π£
2 )π‘
π£=π’+ππ‘π£βπ’=ππ‘π£βπ’π =π‘
π =(π’+π£2 )π‘
π =(π’+π£2 )
π =(π£2βπ’2
2π )2ππ =π£2βπ’2
π’2+2ππ =π£2
π£2=π’2+2ππ
π£2=π’2+2ππ
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!
(π£βπ’π )
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different combination of
βSUVATβ, for a particle moving in a straight line with constant
acceleration
2B
π£=π’+ππ‘π =(π’+π£
2 )π‘π£2=π’2+2ππ
π =(π’+π£2 )π‘
π =(π’+π’+ππ‘2 )π‘
π =( 2π’+ππ‘2 )π‘
π =(π’+ 12ππ‘ )π‘
π =π’π‘+ 12 ππ‘2
Replace βvβ with βu + atβ
Group terms on the numerator
Divide the numerator by 2
Multiply out the bracket
π =π’π‘+ 12 ππ‘2
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different combination of
βSUVATβ, for a particle moving in a straight line with constant
acceleration
2B
π£=π’+ππ‘π =(π’+π£
2 )π‘π£2=π’2+2ππ
π =π’π‘+ 12 ππ‘2
π£=π’+ππ‘π£βππ‘=π’
π =π’π‘+ 12 ππ‘2
π =(π£βππ‘)π‘+ 12 ππ‘2
π =π£π‘βππ‘2+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
π =π£π‘β 12 ππ‘2
Subtract βatβ
Replace βuβ with βv - atβ from aboveβ
Multiply out the bracket
Group up the at2 terms
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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
π =?π’=3 π=5 π‘=?π£=18Write out βsuvatβ
with the information
given
π£2=π’2+2ππ 182=32+2(5)π
324=9+10 π 315=10 π
31.5π=π
Replace v, u and a
Work out terms
Subtract 9
Divide by 10
We are calculating s,
using v, u and a
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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
π =?π’=8 π=0.75π‘=12π£=?
Draw a diagram
Write out βsuvatβ with the
information given
We are calculating s,
using u, a and tπ =π’π‘+ 12 ππ‘2
π =(8Γ12)+ 12 (0.75Γ122)
π =150π
Replace u, a and t
Calculate
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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 β 150mb) The speed with which the car passes the lamp post
8ms-1
Pillar Box
Lamp Post
π =?π’=8 π=0.75π‘=12π£=?
Draw a diagram
Write out βsuvatβ with the
information given
We are calculating v,
using u, a and tReplace u, a
and t
Calculate
π£=π’+ππ‘π£=8+(0.75Γ12)
π£=17ππ β1
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!
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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 Aπ =20π’=13 π=β4π‘=?π£=?
Draw a diagram
Write out βsuvatβ with the
information given
We are calculating t,
using s, u and aπ =π’π‘+ 12 ππ‘2
20=(13)π‘+ 12 (β4 )π‘2
20=13 π‘β2π‘ 2
2 π‘2β13 π‘+20=0(2 π‘β5)(π‘β4 )=0
π‘=2.5ππ 4
Replace s, u and a
Simplify terms
Rearrange and set equal to 0
Factorise (or use the quadratic formulaβ¦)
We have 2 answers. As the acceleration is negative, the particle passes through A, then changes direction and
passes through it again!
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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 β 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 Aπ =20π’=13 π=β4π‘=?π£=?
Draw a diagram
Write out βsuvatβ with the
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 secondsβ¦
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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 β 2.5 and 4 secondsb) The total time the particle is beyond A β 1.5 secondsc) The time taken for the particle to return to O
13ms-1
O Aπ =20π’=13 π=β4π‘=?π£=?
Draw a diagram
Write out βsuvatβ with the
information given
The particle returns to O when s = 0
π =0
π =π’π‘+ 12 ππ‘2
0=(13)π‘+(β2)π‘ 2
0=13 π‘β2π‘ 2
2 π‘2β13 π‘=0π‘ (2π‘β13)=0
Replace s, u and a
π‘=0ππ 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
Kinematics of a Particle moving in a Straight Line
You can also use 3 more formulae linking different
combination of βSUVATβ, for a particle moving in a straight
line with constant acceleration
2B
π£=π’+ππ‘ π =(π’+π£2 )π‘
π£2=π’2+2ππ π =π’π‘+ 12 ππ‘2
π =π£π‘β 12 ππ‘2
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 0π =? π’=15 π=β2.5π‘=?π£=0
π£2=π’2+2ππ 02=152+2(β2.5) π 0=225β5 π 5 π =225π =45π
Replace v, u and a
Simplify
Add 5s
Divide by 5
πππ‘ππ π =90π45m is the distance from O to X. Double it for the total distance travelled
We are calculating s,
using u, v and a