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### Motion - Linear.

posted Feb 10, 2015, 3:45 AM by ranmini@charliesresearch.com   [ updated Jun 20, 2021, 1:29 PM by Upali Salpadoru ]
Speed, Velocity, Acceleration

Speed.

Every motor car                       has a Speedometer.                 It generally gives the speed  in Kilometers Per Hour abbreviated  as KMPH.

In this case the road police will definitely charge the driver.

Fig 1. Car speedometer and a road sign.

Do you know why?

According to the dial the car is moving over 70 km in one hour. The road sign shows  60 km/h which is the maximum speed allowed..

The driver is violating a traffic rule.

What  should the driver do ?                 Accelerate,  Brake or Continue. ?

The driver has two options. He may release the accelerator or he can apply the brakes.

After a journey or some kind of movement there are two quantities that could have been measured. One is the distance, and the other is the time taken.

If a car can maintain a regular speed, the speed multiplied by the time will give give you the distance the car can travel during that time.

 Distance = Average Speed x Time.

We usually divide the distance by the time and say we have obtained the speed.

If the start was zero speed, and the speed had been varying throughout the journey, What speed do we get when we divide the distance by the time taken?

You may be right

.       What we get is the Average Speed.

Average speed = Distance / Time.

 Can we measure the speed of a moving object at any instance?          Yes! We can.  This is what a speedometer of a car does.       But can we get the speed of a moving vehicle from outside?       To get the speed you must know two things. The distance it moves and the time it takes.               These two factors cannot be measured at the same time. So one may say this is impossible.       But    do you know  that the police is doing this all the time.      They employ a completely a different technique using a LASER gun.

Distance and Displacement

In measuring the distance the direction is not important.

Displacement is a quantity that shows how much an object has got displaced in a particular direction.

The ladder is 5 m. long placed against a wall 3 m.  high.

Look at the figure and you will see when the man climbs the ladder he would walk a distance of 5. m.

But how much can  he climb?

Only …3.m.

That shows his vertical displacement is only 3 m.

Displacement can be measured in any other direction too.

For example what is man’s Horizontal displacement?              Only 4 m.

Velocity

Fig.3 Vertical and Horizontal Displacements.

Velocity measures the rate of change of displacement instead of a distance.

In other words velocity has to be only in a particular direction whereas for speed direction may change.

 The quantities where the direction is important are called vectors while the other quantities are called scalars. Vectors= Weight, Velocity, Force.    Scalars= Mass, Speed, Pressure.

Acceleration.

What happens when you press the accelerator?        Speed increases.

A measurement of this change in speed is acceleration.

Acceleration = (Second velocity - First velocity ) divided by Time.

A = V2 - V1 / t.

While velocity is the rate of change in distance, the acceleration is the rate of change in velocity.

Example .1.

A man runs 100 m, starting from rest toward East regularly increasing his speed  taking 10 seconds.

Find the following:-
i.     The average velocity.    ii.    His highest velocity.

iii.    His rate of increasing velocity. Acceleration.

Solution.

i.   Average velocity.

V av = Displacement  divided by Time.

= 100 m /  10 s

.
= 10 ms-1

ii    Final velocity.

If the rate of increasing velocity is uniform you can get the average      velocity by dividing the sum of initial velocity and the final velocity by two.        V1 Initial velocity.      V2  Final velocity.

Therefore   V2  = 2 x V av

V av  =   (V1  +  V2) / 2

V av  =   (V1  +  V2)

2
V2   =  10 x 2s = 20 ms-1

iii.  Acceleration.

The increase of speed is from zero to 20 ms-1.
The time taken for this was 10 s.

Acceleration=  Change of velocity /  Time.

Acceleration      =  (20 ms-1  - 0 )  /  10 s.

Acceleration      =  2 ms-2

Fig. 4   Uniform velocity graphs.

 Graphs Displacement / Time gra Velocity / Tim Shape A straight line showing an upward slope. A straight line parallel to the time axis. Gradient Gr = d/t    = 50 m / 5 s                    = 10 ms1 Zero Displacement Height at any time gives the displacement. The area below the graph. ( This part is coloured green to make it clear)

Solution.

Fig.5  A Graph showing Acceleration.

Make your Comments on the two Graphs A and B and compare with what is given.

 Graph A Graph B i.         Not a straight line.  The shape is        an upward  curve.ii.        The gradient of the graph is                changing. This increase is due to        an increase in velocity. i.        A straight line inclined up.    ii.      The area below the graph gives the                distance traveled.    iii.     Area of green triangle.           Area = ½ base x height.           A=1/2 ( 5x10)             =50. M2( Although the area unit is square meters here the answer has to be in meters.

Formulae

Symbols:- Average velocity = V. av.   First velocity  V 1 ,

Second velocity= V.2     Time = t.,     Displacement = d,         Acceleration=a

 V = d  / t. V.av = (Vi + V.2) / 2          for uniform velocity. V.f = Vi +  a.t a  =   (V 2 – Vi )  /  t. d = Vit + ½ at2

Multiple Choice Questions.

1.0     Which of these will give a vector quantity?

A- Distance / Time.    B- Displacement /Time.   C- Length X Breadth.  D-  Change in speed/Time.

2.0    Which is   the correct definition for acceleration?

A-  Velocity/ time.   B-  Speed / Time.   C- Change in velocity / time taken.   D- displacement/ Time.

The graph  for the  questions from  3.0.  to  5.0

3.0   The graph shows,

A-  the rate of change of speed.  B-  the rate of change of distance,

C-  increasing speed.     D- A distance equal to the area below graph.

4.0    What is the speed during the first 3 seconds?

A- 10 ms-1 ,        B-  90  ms -1       C-  5 ms-1 .        D-  10 ms -2

5.0    What is the speed during the last 2 seconds?

A- 10 ms-1 ,        B-  90  ms -1       C-  5 ms-1 .        D-  10 ms -2

The graph for the questions 6.0 to 9.0.

Graph for questions 6 to 9.

6.0 What is the acceleration during the first 5 seconds?

A - (45 -  20) / 5 ms-2.    B- 45 /5 ms-2.

C- 0 ms-2.    D- 1/2 X 5 X 25 ms-2

7.0  What is the displacement  during the first 5 seconds?

A - (45  20) / 5 m.    B- 45 /5 m.

C- 5 x 45  m.     D- 1/2 X 5 X 25 ms-2

8.0  What is the average velocity?

A. 45 /5 ms-1   B .  45 - 20 /5 ms-1   C. 45 /2 ms-1  D.  (45 + 2o) /2

9.0  What is the correct statement about he graph?

A- Shows regular velocity.                B- Shows regular acceleration.

C-  Shows increasing acceleration.    D- Shows decreasing acceleration.

10.0 An object is dropped and it  falls to the ground for 6 seconds. Taking the acceleration of gravity as 9.8 ms-2 , what is the distance it has fallen?  ( Neglect air friction)

A-    d =  ½ 9.8 x 6x6. m.   B-  6 x 9.8 / 6  m.  C-  9.8 x2  x 6 m. D - 9.8./2 x 6 m .

1, B,   2. C,  3. B,  4. A, 5. A   6. A.  7. D. 8. D  9. B 10.

General questioons.

Q.1.0

A train 50 m. long passes a signal post in 2 seconds. Find the following:
i.      The velocity (speed) of the train.

ii.      What is the time it will take to pass a platform 30 m long?

iiii.    If the train comes to rest in 5 seconds the (negative) acceleration
on braking.     (deceleration)

Q.2.0

A plane bypassing an aerodrome,  cruising  at  150 m.s-1  was observed for 5 s.
On a command by the control tower it reduced the speed (velocity) and landed in 10 seconds.

i.     What would be the displacement at cruising speed?
ii.    What is the negative acceleration in landing?
iii.   What could be the displacement in stopping?

Q.3.0

The downward velocity of a bungee jumper has been recorded and plotted as shown.

i. What is the maximum velocity attained?    ii.     Find the positive acceleration ..        iii.    Find the acceleration during the reducing speed.   iv.    Find the height freely fallen before the chord tightens.

Q. 4.0

The braking system of a car is capable of a deceleration   at the rate of 6 m.s-2
Find the stopping distance when traveling at 30 ms-1

Q.5.0

Examine the Velocity / Time grpha here and answer the questions below

i.   Find  the average velocity for each second separately.

 Time 1 2 3 4 Velocity avg.

II. Find  the distance traveled .

 Time One second Two seconds. Three second Four seconds Displacement

iii.                      Draw a displacement /Time graph.

 Time / s 1 2 3 4 Displacement

Q.1.0

 i. Velocity = distance/ timeV = 50/ 2V= 25 ms-1. ii.Time = 50 + 30 /25          =   80/25           = 3.2 s. iii.  a = Change in velocity / time          = 25 /5            =  - 5 m.s-2
 i. Velocity = distance/ timeV = 50/ 2V= 25 ms-1. ii.Time = 50 + 30 /25          =   80/25           = 3.2 s. iii.  a = Change in velocity / time          = 25 /5            =  - 5 m.s-2

Q.2.0

 i.            V = d  / t.    150 = d x 5  ∴  d   = 150 x 5           =    750 m. ii.  a  =(V f – Vi )  /  t.     a  =(0 - 150) / 10         = -15 m.s-2 iii.   method 1.d = V1t + ½ at2   d = (150 x 10)  + -15x100 d=  1500 -1500/2 =     =  1500 - 750  = 750 mMethod 2.   V.av = (Vi + V.f) / 2            = 75 m/s   d=  V xt     =  750 m.

### i. What is the maximum velocity attained?   ii.     Find the positive acceleration ..        iii.    Find the acceleration during the reducing speed.   iv.    Find the height freely fallen before the chord tightens.

 i.40 ms-1 ii.a =(V 2 – Vi )  /  t.   = (40 - 0) / 4.5   = 8.9 ms-2 iii.a =(V 2 – Vi )  /  t    = (0- 40) /4.5    = - 8.9 ms-2 iv. d =V.av x t V.av = 1/2  (V f +Vi )         = 1/2 (40 + 0)        = 20 ms-1  d =V.av x t      = 20 x 4.5      =4.4 m.