copyright © 2015 chris j jewell 1 mechanics m1 (slide set 4) linear motion (constant acceleration)...
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Copyright © 2015 Chris J Jewell1
Mechanics M1(Slide Set 4)
Linear Motion (Constant Acceleration)
Mechanics M1
Copyright © 2015 Chris J Jewell2
Distance versus Displacement
0
Distance
Displacement
Distance is a scalar
Displacement is a vector
θ Time
DistanceSpeed
Time
ntDisplacemeVelocity Average
Mechanics M1
Copyright © 2015 Chris J Jewell3
Distance versus Displacement
Time
DistanceSpeed Time
ntDisplacemeVelocity Average
0
Displacement
Displacement is a vector
time
10 km
1hr 2hr 3hr
Start Stop+ Vel. -Vel.
Resting
From the start velocity is positive for 1 hour at v = 10 km hr-1, then zero and finally back to the start location again at v = -10 km hr-1
We get the same magnitude but no direction information, so we have speed not velocity
Mechanics M1
Distance
Distance is a scalar
0 1hr 2hr 3hr
10 km
20 km
Resting
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Velocity-Time Graphs
Area A1 + A2 + A3 = Distance Travelled
2850m
60102
1 60536015
2
1Distance
Area under the acceleration graphs represent change in Velocity
A1 = 4 x 15 = 60 ms-1
A3 = - 6 x 10 = - 60 ms-1
Acceleration -Time Graphs
0
60
15
v ms-1
60t sec
a ms-2
t sec
30 45
0 15 6030 45
4
-6
A1
A2
A3
A1
A3
Constant velocity = Zero Acceleration
Mechanics M1
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Motion Graph Examples
Example 1: A bus starts from stop A and accelerates for 20 minutes at 2.5 mmin-2 until it reaches a constant velocity of 50 mmin-1 After 30 minutes it decelerates at 2 mmin-2 to a stop at B. Draw a velocity – time graph of the motion, calculate the time of arrival and the distance travelled by the train.
In triangle A3
min75t1502t
501002t
s 50)(t
50msms 2
t
va
BB
B
B
12
Total distance travelled by the bus is Area A1 + A2 + A3
m 2625(50)(25)2
1(30)(50)(20)(50)
2
1 travelledDistance
(b)
(c)
Mechanics M1
Velocity-Time Graphs
0
50
20
v ms-1
50t min
30 45
A1
A2
A3
tB
BA
(a)
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Motion Graph
Example 2 - A racing car passes the pits at a steady 60 ms-1 and a second car, initially at rest, joins the track with a constant acceleration of 25 ms-2 for 3 seconds until it reaches a constant velocity of 75 ms-1. Calculate the time the second car catches the first car and the distance travelled at that time.
The cars are in the same location when the graph area for each car is the same
7560
v ms-1
t sec0 T3
That occurs when
7.5secT15T112.560T75T112.5225
22575T112.560T753)(T7532
1T60
Total distance travelled is then 60 ms-1 x 7.5 sec = 450 m
Mechanics M1
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Equations of Motion
atuv
t
uva
Since this is a velocity – time graph we can also find the distance travelled in the time interval from the graph
Distance travelled s = the area under the line = the area of the trapezium ABCD
This can be split into the square ABED and triangle DEC to give:
tvus )(2
1
tuvuts )(2
1
uvat since at2
1utsor 2
0
u
v
Velocity
Time
Final Velocity
Initial Velocity
Time Interval (t)
interval Time
velocityInitial velocityFinalionacccelerat
tuvs )(2
1
tA tB
A B
D
C
E
uts
Mechanics M1
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(4) 2asu v
u)v)(v(u2asa
uv
2
vu s
get weabove (2) and (1) from t eliminate weIf
22
(3) at2
1- vt sor at
2
1ut s
(2) v)t (u2
1s
(1) at uv
22
SUMMARY
Where “a” is acceleration (ms-2), “s” is the displacement (m), “t” is time (s), “u” is initial velocity (ms-1) and “v” is the final velocity (ms-1) – these are also referred to as “SUVAT” equations.
Newton’s Laws of Motion
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Example 1: A particle starts at a point when t = 0, passes point A after 5 sec and point B after 6.8 sec, if the car has a constant acceleration 1.5 ms-2 . If the initial speed of the car is 10 ms-1, find (a) the distance AB and (b) the velocity after 10 sec?
In this case we have u = 10 ms-1, v = ? ms-1 a = ? ms-2 , Time from the start to point A is 5 seconds
atu vUsing
1-
-1
ms 20.2 6.81.5 10 vBAt
ms 17.5 51.510A vAt
m 33.9 AB Distance
8.12
2.205.17 st
2
uv s Using
A B
(a) (b) atu vUsing
1-ms 25 101.510v
seconds 10After
Mechanics M1
Straight Line Motion Examples
t = 0 t = 5 sec t = 6.8 sec
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Example 2: A particle passes point A with a speed of 15 ms-1 and point B with a speed of 30 ms-1. If the particle has a constant acceleration, find the speed of the particle at the midway point between A and B.
A BM
(i) ........ 2as1530 2asuv 2222 Moving towards the right, using
(ii) ........ 2as3015 2asuv 2222 Moving towards the left, using
Adding equations (i) and (ii) eliminates 2as
1M
222M
2M
222M
23.7ms562.52
225900 v
2as2as15302v2as15 v 2as30v
Mechanics M1
Positive
15 ms-1 30 ms-1
Copyright © 2015 Chris J Jewell11
Examination Type Questions
A train moves with constant acceleration along a straight horizontal track. It passes point A with a speed of 6 ms-1 and 10 seconds later it passes point B. If AB =100 m find (a) the acceleration of the train (b) the speed of the train when it passes point C which is 120 m further down the track,.
A
6 ms -1
B
? ms -1
100 m C120 m
? ms -1
We know u = 6 ms-1 at A , v = ?, t = 10 sec and AB = s = 100 m.
2-
1
ms 0.8a 10a614atuin v Substitute
14ms v 1406020010v10v60200
10v)(62
1100v)t(u
2
1s
(a) (b)12
2222
19.7msv388v
1208.0214v2asuv
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Copyright © 2015 Chris J Jewell12
M1 Mechanics
Vertical Motion
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Copyright © 2015 Chris J Jewell13
Mechanics M1
Vertical motion under gravity. The equations of motion previously described can be applied directly to vertical motion with the acceleration term being always the value due to gravity.
It is very important to realise that the equations of motion can be used to evaluate positive and negative values of velocity and displacement, as long as the direction of motion is specified relative to a reference level: Positive up, Negative down.
Applications of the velocity formula v = u + at can be illustrated by considering a particle being projected vertically upward with a velocity of 30 m s-1, say, and plotting the results as a velocity – time graph (assuming g = -10 m s-2 for convenience)
The value of v = u + gt is shown opposite (red spheres) for t = 0 to t = 6 seconds, defining positive as upward. The initial velocity of u = 30 m s-1 is gradually reduced to v = 0 after 3 s, then increases to – 30 m s-1 at time t = 6 s.
Continued on next slide
0 1 2 54
20
-10
-20
-30
6
v (m s-1)
t (s)
30
3
Velocity -Time Graph
45 m10
- 45 mPositive
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Mechanics M1
Continued on next slide
(1) As the particle rises its velocity decreases to zero – which occurs at its maximum height above its start point:
20s 900s102 - 30 02as- u v:height Maximum (1) 222
(2) When the particle returns to its start point its speed is the same as its initial speed, but in the opposite direction – the motion is symmetrical.
s 3 t (3) The total distance of the flight is twice distance to the highest point.
Vertical motion under gravity.
Notes:
height maximum 2 : travelleddistance Total (3)
m 45s
10t - 30 0at u v:height maximumreach toTime (2)
m 90height Maximum
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Mechanics M1
Continued on next slide
2s m 106
60- isgraph t - v theof slope The (5)
m 90 of totala giving s 6 and s 3between m 45 - and s 3 and s 0between m 45:comprisesgraph under the area The (6)
These considerations are analysed in more detail below related to the particle projected at 30 m s-1:
Vertical motion under gravity.
(5) The slope of the graph (on the previous slide) is – 10 m s-2
(6) The area under the velocity-time graph gives the total distance travelled.
(4) Total travel time can be found using v = u + at, with final velocity = - u
t10 6010t 2u at u - v: time travelTotal (4)
s 6t
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Mechanics M1
Displacement
The formula for displacement can be written as:
This equation completely describes the upward and downward motion of a body under the effect of gravity near the surface of the Earth (excluding any other effects such as air resistance). The nature of this quadratic equation is illustrated with a displacement - time graph below, g has been taken as -10 m s-1:
2gt2
1uts
Note that the axes are displacement and time – But the particle motion is vertically up and down.
The motion is symmetrical with the particle exactly repeating each position.
Continued on next slide
30
1 2 54
40
50
20
10
t (s)
60
3 60
s (m) Displacement-Time Graph
Positive
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Mechanics M1
Calculation Examples
(a) To find out the time when the particle returns to the ground the equation is used as follows:
6or t 0tt)5t(6
(b) To find the time when the particle, projected from the ground is at 25 m from the start position (+ 25 m above the ground) the equation is used as follows:
2530t5t05t30t25gt2
1uts 222
30
1 2 54
40
50
20
10
t (s)
60
3 60
s (m) Displacement-Time Graph
22 5t30t0gt2
1uts
s 5 tand s 1t05)1)(t(t056tt 2
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Mechanics M1
Calculation Examples
(c) To find out the time that the particle hits the ground if the particle’s start position is 25m above the ground the equation is used as follows:
30
1 2 54
40
50
20
10
t (s)
60
3 60
s (m) Displacement-Time Graph
25 mParticle thrown from here (Reference Zero)
-0.74 6.74
D – T Curve for Particle thrown from 25 m above the ground
2
51466
2a
4acbbt
22
056tt5t-30t2505t30t-25gt2
1uts 2222
7417.02
7.4833-6 or t 6.74
2
748336t
Displacement to ground is – 25 m
Positive
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Mechanics M1
Further Example Calculations
A ball projected upwards with a velocity of 13 m s-1 from a height x above the ground – Neglecting air resistance work out x if the ball takes 3 s to reach the ground.
Neglecting air resistance work out the velocity of the ball when it hits the ground.
39.813vatuv:thenpositive, as up Define
Reference Zero
m 5.1x
22 34.9313xgt2
1uts
:thenpositive, as up Define
1s m 16.4v - x
13 m s -1
16.4 m s -1
Positive
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Motion under Gravity – Acceleration is always downwards at 9.8 ms-2
Example 1 : A stone is dropped into a well and a splash sound is heard 3.5 seconds later, how deep is the well?
We have t = 3.5 sec, u = 0 ms-1, v = 0 ms-1 a = 9.8 ms-2 and we need to find s,, so we can use:
s
at2
1ut s 2
(3.5)9.82
1(3.5)0 s 2
m 60 s
Mechanics M1
Positive
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Example 2 : A tennis ball thrown vertically from a window 32 m above the ground with a velocity of 16 ms-1, find (a) the maximum height above the ground attained by the ball (b) the time that the ball touches the ground?
(a) Maximum height: From v2 = u2 + 2as, since v = 0 at the highest point:
13.1m19.6
256s19.6s256or
s(-9.8)2 160as2u v 222
1.50secor sec 4.56x4.92
32)(4.94(16)(-16)x
2
4bb- xformula quadratic theUsing
2
2
a
ac
-32 m
+ve
So the maximum height is 32 m + 13.1 m = 41.5 m
Not to scale:
032164.9t
4.9t-16t-32t(-9.8)2
1t16 32-
2
22
tZero Level
Maximum Height
+13.1 m
Mechanics M1
(b) Time to reach the ground
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Examination Type Questions
A ball is projected upwards from the ground with a velocity of 24 ms-1. Find (a) its maximum height (b) the times that the ball is more than 24 metre above the ground.
(a) Using v2 = u2 + 2as the highest point is reached when v = 0 i.e. when 0 = 576 – 19.6s or s = 576 / 19.6 = 29.4 m
Maximum Height
24 m
Ground
(b)
sec 3.5 and 1.40 t 9.42
249.44)24()24(
02424t4.9t4.9t24t24
9.8msg and,ms 24u m, 24sset we
gt2
1uts Using
2
2
2
2
1
2
t
(c) Time above 25 m = 3.5 – 1.4 = 2.1 sec
Mechanics M1