motion

1. MOTION IN A STRAIGT LINE
An object is said to be in motion when its position is continuously changing relative to a
reference point such as observer or a detection device.
Linear motion is a movement in a straight line.
Distance and displacement
Distance is physical quantity which specifies magnitude only.
Displacement is the physical quantity which specifies magnitude and direction.
Speed and velocity
Speed is the distance travelled per unit time.
Speed has magnitude only.
 𝑠𝑝𝑒𝑒𝑑(𝑣) =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑑)
𝑡𝑖𝑚𝑒(𝑡)
 𝑣 =
𝑑
𝑡
 The S.I unit of speed is m/s
Velocity is the distance travelled in a specific direction per unit time.
 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦(𝑣) =
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡(𝑠)
𝑡𝑖𝑚𝑒(𝑡)
 𝑣 =
𝑠
𝑡
 The S.I unit of velocity is m/s
A body is said to move with uniform velocity if its rate of change of displacement with time is
constant.
Example 01.
An object travelled 20m to the right in 4s and then 12m to the left in 3m. for its total motion.
What its average speed and its average velocity.
Soln
Data given
Distance to the right (d1)=20m
Distanceto the left (d2)=12m
Total distance(d) =d1+d2=20m+12m=32m
Displacement to the right (s1)=20m
Displacement to the left (s2)=-12m
Total displacement(s) =s1+s2=20m+(-12m)=8m

2. Note: distance and time alwas is positive;
Total time(t)=4s+3s=7s
Formula
(a)Speed,v=d/t
(b )Velocity,v=s/t
Calculation
a). v =
32m
7s
= 4.57m/s
b) 𝑣 =
8𝑚
7𝑠
= 1.14𝑚/𝑠
Acceleration and deceleration/retadation
Acceleration is defined as the rate of change of velocity.
Deceleration/retardation is the negative rate of change of velocity.
Or
Is the negative acceleration.
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛( 𝑎) =
𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦(𝑣) − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦(𝑢)
𝑡𝑖𝑚𝑒(𝑡)
𝑎 =
𝑣 − 𝑢
𝑡
𝑇ℎ𝑒 𝑆. 𝐼 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 m/𝑠2
Example 01
An object is initially moving at 15m/s to the right. Eight seconds later it is moving at 5m/s to
the left. During those eight seconds, what was the object’s acceleration?
Soln
Data given
Initial velocity(u)=15m/s
Final velocity(v)=-5m/s
Time taken(t)=8s
Required: acceleration(a)=?
Formula
𝑎 =
𝑣 − 𝑢
𝑡

3. Calculation
𝑎 =
−5 − 5
8
= −
20
8
= −2.5𝑚/𝑠2
Therefore the object’s acceleration is 2.5m/𝑠2
The negative sign indicates that the direction of acceleration was to the left;
Example 02
A car brakes and slows down from 20m/s to 5m/s in 3 seconds. What is the vehicle’s
acceleration?
Soln
Data given
Initial velocity(u)=20m/s
Final velocity(v)=5m/s
Time taken(t) =3s
Required: acceleration(a)=?
Formula
𝑎 =
𝑣 − 𝑢
𝑡
Calculation
𝑎 =
5 − 20
3
=
−15
3
= −5𝑚/𝑠2
This means the car decelerate or slows down by 5𝑚/𝑠2
Therefore its retardation is 5𝑚/𝑠2
Position-time graph
Graphs can be a useful method of presenting data alongside relationships between parameters
such as displacement, velocity acceleration and time.
Distance time graph
On a distance versus time graph, motion at a constant velocity is represented by a straight lines
and the slope of the line represents the velocity of the object.
distance
time

4. Velocity-time graph
The area under velocity time graph represents acceleration