Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 4) Linear Motion (Constant Acceleration)...

Copyright © 2015 Chris J Jewell1

Mechanics M1(Slide Set 4)

Linear Motion (Constant Acceleration)

Mechanics M1


Distance versus Displacement

0

Distance

Displacement

Distance is a scalar

Displacement is a vector

θ Time

DistanceSpeed

Time

ntDisplacemeVelocity Average

Mechanics M1


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


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


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)


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


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


(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

Mechanics M1


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


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


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

Mechanics M1


M1 Mechanics

Vertical Motion

Mechanics M1


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


Mechanics M1


(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


Mechanics M1


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


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.


30

1 2 54

40

50

20

10

t (s)

60

3 60

s (m) Displacement-Time Graph

Positive


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


22 5t30t0gt2

1uts

s 5 tand s 1t05)1)(t(t056tt 2


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


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


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


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


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


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

Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 4) Linear Motion (Constant Acceleration)...

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Transcript of Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 4) Linear Motion (Constant Acceleration)...