Kinematics (a) define displacement, speed, velocity and acceleration. 1. DEFINITIONS OF TERMS Displacement - the linear distance of an object or a point, in a specified direction, from some reference point. Speed - the rate of change of the total distance travelled. Velocity - the rate of change of displacement of a body. Acceleration - the rate of change of velocity of a body. 1.1 Displacement Example 1 A car travels 3.0 km due east from point O to A and then a further 3.0 km due north of A. What are the distance covered and its displacement?Distance, x = 3 + 3 = 6 km s 3 km 3 km Displacement, s =

32 + 32= 4.2 km at 45

1.2 Average speed and average velocity Average speed refers to the total distance travelled over the total time taken. Average speed =

total distance travelled total time taken

Units: m s-1

Average velocity refers to the total displacement travelled over the total time taken.total displacement total time taken

Average velocity =

Units: m s-1

The average speed is not equal to the magnitude of the average velocity.

1.3 Instantaneous speed and average velocity Instantaneous speed is the speed at a particular point or a particular instant of time. Instantaneous speed = gradient of the distance-time graph at that instant Instantaneous velocity is the velocity at a particular point or a particular instant of time.

v = lim

s ds = t 0 t dt

Hence instantaneous velocity v can be defined as rate of change of displacement It is equal to the gradient of the displacement-time graph.

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Instantaneous vs Average Instantaneous ds v= dt dv a= dt Gradient at the point in time 1.4 Acceleration An object whose velocity is changing is said to have an acceleration.

Average s < v >= t v < a >= t Total divided by time

s s1 s (0,0) t t1 t

< v >=

s s1 = t t1

v=

ds dt

t = t1

a = lim

t 0

v dv = t dtUnit: m s-2

It is a vector quantity.

Acceleration can arise from a change in either magnitude or direction of the velocity or both.

Example 2Reference

Point

Object Ambulance Motorcycle Cow Race Car

Sign of Velocity -ve +ve -ve +ve

Sign of Displacement -ve -ve +ve +ve

(b) use graphical methods to represent displacement, speed, velocity and acceleration; (c) find displacement from the area under a velocity-time graph; Velocity-time graph: Area under graph =>displacementv/ms-1

t/s

The area beneath the v-t graph is the displacement. The area can be found by counting

squares if the area is irregular. 2

(d) use the slope of a displacement-time graph to find the velocity; s/m Displacement-time graph: Gradient => velocity

s t t1

t/s

(e) use the slope of a velocity-time graph to find acceleration. Velocity-time graph: Gradient => acceleration

v/ms-1

v t

t1 Acceleration-time graph: Area under graph => velocitya/ms-2

t/s

t/s

The area beneath the a-t graph is the distance.

Graph s-t v-t a-t

Gradient v a -

Area under graph s v

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(f) derive, from the definitions of velocity and acceleration, equations which represent uniformly accelerated motion in a straight line; 2. DERIVATIONS OF EQUATIONS OF MOTION

Let us consider a body experiencing a uniform acceleration, a, while moving in a straight

line. We will consider its motion during a time interval t. Let u be its velocity at the beginning of the internal, v the velocity at the end of the internal

and s its displacement during the interval t. Since we have earlier learnt that acceleration = gradient of v-t graph,

a =

v u change in velocity = change in time t= area under v-t graph,

(1)

We also learnt that displacement

Or for constant a, average velocity = (initial velocity + final velocity)/2 vave = (v + u)/2 hence s = vave t => s=1 2

(v + u )(t )

(2)

Rearranging (1), we obtain the first of three equations as given

I. Substituting I. into (2),

v

=

u + at

s

= II.

1 2

[(u + at ) + u ](t )= ut +1 2

=

1 2

[2u + at ](t )

s

at 2

Combining (1) & (2) and eliminating t,

s

=

1 2

(v + u ) v u a v2 u 2

2as

=

III.

v2 =

u 2 + 2as

Condition for application of equation of motion

1. uniform acceleration 2. linear motion 4

(g) solve problems using equations which represent uniformly accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance. 3. MOTION DUE TO GRAVITY

When the pull of gravity is the only force experienced by an object near the Earth, the

object is said to be in free fall. A body in free fall (Earth) experiences a constant acceleration of

g = 9.81 ms-2 Example 3 A boy standing at the edge of a cliff of 20 m above sea level, throws a stone straight up with a speed of 25 ms-1. (a) How long does it take to reach its highest point? (b) How high does the ball rise above its release point? (c) How long will it take for the stone to reach a point 25 m above its release point? (d) Calculate the velocity of the stone when it hits the water surface. Sign Convention: & positive, (a) v = u + at 0 = 25 + (-9.81)t t = 25 / 9.81 = 2.5 s s = ut +1 2

(b)

at 2

= 25 (2.5) + (0.5)(-9.81)(2.5)2 = 31.8 m (c) s = ut +1 2

at 2

25 = 25t + (0.5)(-9.81)t2 4.905t2 -25t + 25 =0

(25) (25) 2 4(4.905)(25) t= 2(4.905)t = 1.37s or 3.73 s (d)v2 = u 2 + 2as

(3sf)

= (25)2 + 2(-9.81)(-20) v = 31.9 ms-1

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(h) describe qualitatively the motion of bodies falling in a uniform gravitational field with air resistance. 4. EFFECTS OF AIR RESISTANCE

The force that the air exerts on a body moving through it, called air resistance (the

technical term is drag force), always act in a direction opposite to the object's velocity. The magnitude of the air resistance depends on the density of the air and on the speed of

the body through the air; the faster the body a moves, the greater air resistance. When a body is falling vertically from rest, the weight is always constant and acting

downwards throughout. The table below gives the stages of motion that describes how air resistance and

subsequently the body motion changes. Qualitative Description of Motion At the instant which the object is released: A u = 0 ms-1 No air resistance Forces F = mg Diagrams

Net force acting on object = weight W. Object accelerates and downward velocity increases. Downward velocity Upward air resistance. F = mg - Fd F>0

B

Net force acting on object < W Object accelerates and downward velocity increases. Air resistance = Weight Net force acting on object = 0 N

Fd = mg F=0

C

Net Acceleration = 0 ms-2 Velocity does not increase any further (Terminal velocity vt ).

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Consider a body being dropped from rest. (u = 0) Choosing the origin at the point of projection, downwards as the positive direction.

Without air resistancea/ms-2

With air resistancea/ms-2

g

g

0v/ms-1

t

0v/ms-1

t

vt

0s/m

t

0s/m

t

0 Velocity

t

0

t

Free Falling object (under constant gravitational acceleration)

Without air resistance : constant gradient (constant acceleration) => constant rate of increase in velocity With air resistance : (i) decrease in gradient (decrease in net acceleration:

due to increasing air resistance, Rair) => velocity increasing at a decreasing rate (ii) gradient decreases to zero: if time long enough (no net acceleration: Weight = Rair => constant terminal velocity reached Displacement Without air resistance: a symmetrical graph With air resistance: (i)

)

v

w/o Rair

Decrease in maximum height reached and range of object (ii) Graph most likely asymmetrical s w/o Rair

w Rair 7 t

w Rair t

(i) describe and explain motion due to an uniform velocity in one direction and an uniform acceleration in a perpendicular direction. 5. PROJECTILE MOTION

Sometimes, bodies in free fall have an initial horizontal velocity. In such cases, the motion

is no longer occurs in a straight line. The motion is now non-linear (also known as projectile motion). Consider a body projected at an angle and with initial velocity u. Projectile Motion Motion with uniform velocity in one direction (x) and uniform acceleration in a

perpendicular direction (y) x-direction: Uniform velocity => use sx = ux t y-direction: Uniform acceleration => use Equations of Motion (E.O.M)

CASE 1: Object moving up and then down

CASE 2: Object moving only downwards ux = u

uy = u sin

uy = 0

ux = u cos

_x__ ux = u cos sx = ux t => sx = (u cos )t

__y__ uy = u sin use E.O.M e.g. v = u + at vy = uy + (-g)t vy = u sin + (-g)t NOTE: At maximum height, vy = 0

__x__ ux = u sx = ux t => sx = u t

__y__ uy = 0 use E.O.M e.g. s = ut + at2 sy = uy t + gt2

sy = gt2e.g. v = u + at

vy = gt

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