Form 4 Chapter 2 Forces and Motion2.1 Linear Motion1. Linear
motion is motion in a straight line.2. The study of the motion of
an object without considering the forces acting on it is called
kinematics. 3. Under linear motion, we study thea. distance and
displacementb. speed and velocityc. Acceleration and the
relationship between them.DistanceDefinition: The distance traveled
by an object is thetotal lengththat is traveled by that object.SI
unit: meter (m)Quantity:Scalar
DisplacementDefinition:Displacement of an object from a point of
reference, O is theshortest distanceof the object from point O in
aspecific direction.SI unit:meter (m)Quantity:VectorExample:
The distance of point B from the origin O is 100m.The distance
of point A from the origin O is also 100m.
The displacement of point B from the origin O is +100m.The
displacement of point A from the origin O is -100m.The + and sign
show the direction of the displacement.
Example:
Ali go to work by motorcycle everyday as shown in the diagram
above.The distance that Ali travels from his house to the factory
is 200m.The displacement of Ali from his house after arriving at
the factory is 120 m.
Speed1. Speed is defined as therate of changeindistance. It is a
measure of how fast the distance change in a movement.2. Speed is
ascalar quantity.3. The SI unit of speed is m/s (metre per
second)Equation of Speed
Velocity1. Velocity is defined as therate of change in
displacement. It is the measure of how fast the displacement change
of a moving object.2. Velocity is avector quantity.3. The unit of
velocity is m/s (metre per second)Equation of velocity
Acceleration1. Acceleration is defined as therate of change of
velocity. It is a measure ofhow fast the velocity change.2.
Acceleration is avectorquantity.3. The unit of acceleration
isms-2.4. An object moving with avelocitythat isdecreasingis said
to be experiencingdeceleration.Equation of acceleration
Example:A car travels from a stationary position and reach a
velocity of 36 ms-1in 8 seconds. What is the acceleration of the
car?Answer:Initial velocity, u = 0Final velocity, v = 36 ms-1Time
taken, t = 8sAcceleration, a = ?a = = = 4.5 ms-2Equation of Uniform
AccelerationMost of the motion problems can be solved by the
following equations. Therefore, make sure that you memorise all of
them.
How we know when to use the equation?
The 4 equations are used when the motion is uniform
acceleration.
Motion with Uniform Acceleration Example 1: An object
accelerates from stationary with the acceleration of 4 ms-2. What
is the velocity of the object after 7s?
Answer:It's advisable to list down all the information that we
have.
Initial velocity, u = 0 (Because the motion start from
stationary)Acceleration, a = 4 ms-2Time taken, t = 7sFinal
velocity, v = ?
The displacement, s, is not involved, hence we select the
equationv = u + at to solve the problem.v = u +at; v = (0) + (4)
(7); Final velocity, v = 28ms-1Example 2A car is moving with
velocity 5ms-1reaches a velocity of 25ms-1in 5s. What is the
acceleration of the car?
Answer:It's advisable to list down all the information that we
have.
Initial velocity, u =5ms-1Final velocity, v = 25ms-1Time taken,
t = 5sAcceleration, a = ?
Example 3A cyclist riding at a speed of 40 ms-1braked with
uniform acceleration and stopped in 40m. How long did he take to
stop?
Answer:Initial velocity, u =40 ms-1Final velocity, v = 0
(Because the cyclist stop) Displacement, s = 40mTime taken, t =
?
Example 4A car is accelerated at 4 ms-2from an initial velocity
of 5 ms-1for 10 seconds. What is the distance travelled by the
car?Answer:Acceleration, a = 4 ms-2Initial velocity, u = 5Time
taken, t = 10sDisplacement, s = ?
Example 5A car accelerates from 4 ms-1reaches a velocity of 28
ms-1after travelling for 64m. What is the acceleration of the
car?Answer:
Initial velocity, u =4 ms-1Final velocity, v = 28
ms-1Displacement, s = 64mAcceleration, a = ?
Example 6A car begins to move from rest. The velocity of the car
increases at a rate of 4 ms-2. Find the distance travelled by the
car after 12 second.Answer:
Initial velocity, u = 0 (Because the begins to move from
rest)Acceleration, a = 4 ms-2(Rate of increment = Acceleration)Time
taken, t = 12sDisplacement, s = ?
Example 7A body is accelerated uniformly from rest and in the
first 6.0 s of its motion it travels 30 m. Findi. the average speed
for this period of 6 s,ii. the speed at the end of this period,iii.
the acceleration.
Answer:
Initial velocity, u = 0 (Because the motion start from rest)Time
taken, t = 6.0sDisplacement, s = 30mFinal velocity, v =
?Acceleration, a = ?
i. Average speed = = = 5.0 ms-1
ii. the speed at the end of this period iii. the acceleration.s
= (u + v) ts = ut + at230 = (0 + v) 630 = 0 + a (6)2v = 10 ms-118a
= 30a = 1.667 ms-2
Challenging Question 1:A car starts from rest and accelerates at
a constant acceleration of 3 ms-2for 10 seconds. The car then
travels at a constant velocity for 5 seconds. The brakes are then
applied and the car stops in 5 seconds. What is the total distance
travelled by the car?
Challenging Question 2: Ali starts driving his car from home
with a constant acceleration and reaches a velocity of 30 m/s in
6.0 seconds. Finda. The acceleration of Ali's car.b. The
displacement of Ali's car 5.0 seconds after it started moving.c.
The displacement of Ali's car in the fifth second..d. Velocity of
Ali's car at time t = 4.0 seconds?e. Velocity of Ali's car after
moving 30.0 meters from the starting point.2.1 Ticker - Timer
1. A ticker - timer consists of an electrical vibrator which
vibrates 50 times per second.2. This enables it to make 50 dots per
second on a ticker-tape being pulled through it.3. The time
interval between two adjacent dots on the ticker-tape is called one
tick.4. One tick is equal to 1/50 s or 0.02 s.
Example:Find the number of ticks and the time interval between
the first dot and the last dot on each of the ticker tapes below.
The frequency of the ticker timer is equal to 50Hz.a.Answer:Number
of ticks = 15 Time interval = 15 x 0.02s = 0.3s
b.Answer:Number of ticks = 5 Time interval = 5 x 0.02s =
0.1s
c.Answer:Number of ticks = 8 Time interval = 8 x 0.02s =
0.16sThe distance between dots on a ticker tape1. The distance
between two adjacent dots on a ticker-tape represents the
displacement of the object in a tick (0.02 s).2. If the object
moves quickly, the dots are far apart. If the object moves slowly,
the dots are close to each other.
Analysing Ticker Tape
Finding Velocity from Ticker TapeFinding VelocityVelocity of a
motion can be determined by using ticker tape through the following
equation:
Caution! t is time taken from the first dot to the last dot of
the distance measured.Example:
Diagram above shows a strip of ticker tape that was pulled
through a ticker tape timer that vibrated at 50 times a second.
What is the
a. Time taken from the first dot to the last dot?b. Velocity of
the object that is represented by the ticker tape?
Answer:a. There are 15 ticks from the first dot to the last dot,
hence Time taken = 15 0.02s = 0.3sb. Distance travelled =
15cmVelocity = = 50
Caution! t is time taken from the initial velocity to the final
velocity.
Example:
The ticker-tape in figure above was produced by a toy car moving
up a tilted runway. If the ticker-tape timer produced 50 ticks per
second, find the deceleration of the toy car.Answer:Initial
velocity,u = =150 Final velocity,v = = 25 Time taken for the
velocity change,t = (6-1) ticks = 5 tickst = 5 0.02s = 0.1s
a = = = 1250 Deceleration = 1250
2.2 Analysing Motion Graph
2.2 Analysing Motion Graph
Velocity-time Graphs
Example :
Displacement-time Graph
Velocity-time Graphs
Comparison between the displacement-time graph and velocity-time
graph:
Non-uniform motion
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