NAIJA EXAM RUNZ ONLINE PHYSICS TUTORIAL: MOTION
MOTION
Definition:
Motion is defined as the change of position with time. For a body to move from one place to another it required a particular period of time and when this occur we say the body is in motion.
(Waec GCE 2016 English Answers)
Types of Motion:
(1). Random Motion: This is when a body is moving in a nonlinear manner and changes direction continuously. This motion occur when the gas molecules are moving. It also occur when you release small object e.g feather in air.
(2). Linear Motion: This is a motion of an object in a straight direction. When a car is moving, it moves in a linear motion.
(3). Oscillatory Motion: This a to and fro movement of an object about a fixed position. It occur when you punch a hanged bag.
(4). Rotational Motion: The movement of a body in a rotational manner about it axis. You can see this when the fan blades are rotating about an axis.
EQUATIONS OF A UNIFORMLY ACCELERATED LINEAR MOTION
1st equation of motion; v=u+at
2nd equation of motion: s=t(v+u)/2
3rd equation of motion: s=ut+1/2at^2
4th equation of motion: v^2=u^2+2as
Where: u=initial velocity of the
body(m/s).
v=final velocity of the
body(m/s)
s=distance moved by the
body(m)
t=time taken during the
motion(s)
a=acceleration of the
body(m/s^2)
For a body in acceleration, all the above equations works. When the acceleration is decreasing, the positive value in the equation changes to negative and this is called RETARDATION/DECELERATION..
1st equation of motion; v=u-at
2nd equation of motion: s=t(v-u)/2
3rd equation of motion: s=ut-1/2at^2
4th equation of motion: v^2=u^2-2as
For a body from rest, u=0. Therefore, the equations becomes:
1st equation of motion; v=at
3rd equation of motion: s=1/2at^2
4th equation of motion: v^2=2as
For a body at rest, v=0 and the equations becomes;
1st equation of motion; 0=u+at
4th equation of motion: 0=u^2+2as
Applications Of Equations Of
Motion
1. Make sure that all the units match i.e v in m/s, t in sec, s in m, a in m/s^2.
2. The variables (v,u,t,a,s) in each equations are 4/5 and you are always given three of them required to find one or both unknown.
3. The best way to choose the equation to use is to find out which of the five variables are given and is/are to be determined by first write the data given in the question.
Then find the equation which does not contain the variable that is not given but contain the one to be determined.
SPEED UNITS CONVERSION:
1 Km/hr= (1000/60×60)m/s
72km/hr=72×1000/60×60=20m/s.
1m/s=60×60/1000km/hr.
30m/s=30×60×60/1000=108km/hr.
Worked Example:
1. A body accelerates uniformly from rest at the rate of 3m/s^2 for 8s. Calculate the distance covered by the body during the acceleration NECO.
Solution:
Given that: a=3m/s^2. t=8s, s=?.
Apply; s=1/2at^2 (since the body is from rest).
S=1/2×3×8^2=1/2×3×64=96m.
2. The motion of an object moving in a straight line is uniformly retarded so that it comes to rest in 20s after travelling a distance of 40m. What is the initial velocity of the body? SSCE.
Solution:
Data given; t=20s, s=40m, u=?, v=0 (body at rest.
S=t(v+u)/2
40=20(0+u)/2
80=0+20u (cross multiply).
U=80/20
U=4m/s.
MOTION OF A FREE FALL OR
MOTION UNDER GRAVITY
Acceleration 'a' is replaced by acceleration due to gravity 'g' when dealing with a body falling to the earth.
So anywhere you see 'a' in any of the equations of motion will change to 'g'. i.e
1st equation of motion; v=u+gt
3rd equation of motion: s=ut+1/2gt^2
4th equation of motion: v^2=u^2+2gs
Note:
* For all body moving upward, g is negative and v=0 i.e
1st equation of motion; v=u-gt
3rd equation of motion: s=ut-1/2gt^2
4th equation of motion: v^2=u^2-2gs
* If the body is released from height, u= 0 and s= h. From s=1/2at^2, therefore we have h=1/2gt^2..
* Provided that there is no air resistance, two bodies of different masses get to the ground at the same time. This obey the equation of time of flight which says:
t=√2h/g.
The equation reveals that their time of flight is independent on their masses.
Worked Example:
A palm fruit drops to the ground from the top of a palm tree 80m tall. How long does it take to reach the ground? (g=10m/s^2).
Solution:
Given that; h=80m, g=10m/s^2, t=?.
From h=1/2gt^2.
t=√2h/g=√2×80/10=√160/10=√16=4
t=4s.
Exercise:
Two grapefruit of masses 50g and 70g respectively are falling from the top of the tree simultaneously. Given that the height of the tree is 32m, calculate the time taken by the heavier fruit to hit the ground. Neglect air resistance. (g=10m/s^2)..
END OF LECTURE.
IS THIS HELPFUL?
Tell us if you find this helpful.
Definition:
Motion is defined as the change of position with time. For a body to move from one place to another it required a particular period of time and when this occur we say the body is in motion.
(Waec GCE 2016 English Answers)
Types of Motion:
(1). Random Motion: This is when a body is moving in a nonlinear manner and changes direction continuously. This motion occur when the gas molecules are moving. It also occur when you release small object e.g feather in air.
(2). Linear Motion: This is a motion of an object in a straight direction. When a car is moving, it moves in a linear motion.
(3). Oscillatory Motion: This a to and fro movement of an object about a fixed position. It occur when you punch a hanged bag.
(4). Rotational Motion: The movement of a body in a rotational manner about it axis. You can see this when the fan blades are rotating about an axis.
EQUATIONS OF A UNIFORMLY ACCELERATED LINEAR MOTION
1st equation of motion; v=u+at
2nd equation of motion: s=t(v+u)/2
3rd equation of motion: s=ut+1/2at^2
4th equation of motion: v^2=u^2+2as
Where: u=initial velocity of the
body(m/s).
v=final velocity of the
body(m/s)
s=distance moved by the
body(m)
t=time taken during the
motion(s)
a=acceleration of the
body(m/s^2)
For a body in acceleration, all the above equations works. When the acceleration is decreasing, the positive value in the equation changes to negative and this is called RETARDATION/DECELERATION..
1st equation of motion; v=u-at
2nd equation of motion: s=t(v-u)/2
3rd equation of motion: s=ut-1/2at^2
4th equation of motion: v^2=u^2-2as
For a body from rest, u=0. Therefore, the equations becomes:
1st equation of motion; v=at
3rd equation of motion: s=1/2at^2
4th equation of motion: v^2=2as
For a body at rest, v=0 and the equations becomes;
1st equation of motion; 0=u+at
4th equation of motion: 0=u^2+2as
Applications Of Equations Of
Motion
1. Make sure that all the units match i.e v in m/s, t in sec, s in m, a in m/s^2.
2. The variables (v,u,t,a,s) in each equations are 4/5 and you are always given three of them required to find one or both unknown.
3. The best way to choose the equation to use is to find out which of the five variables are given and is/are to be determined by first write the data given in the question.
Then find the equation which does not contain the variable that is not given but contain the one to be determined.
SPEED UNITS CONVERSION:
1 Km/hr= (1000/60×60)m/s
72km/hr=72×1000/60×60=20m/s.
1m/s=60×60/1000km/hr.
30m/s=30×60×60/1000=108km/hr.
Worked Example:
1. A body accelerates uniformly from rest at the rate of 3m/s^2 for 8s. Calculate the distance covered by the body during the acceleration NECO.
Solution:
Given that: a=3m/s^2. t=8s, s=?.
Apply; s=1/2at^2 (since the body is from rest).
S=1/2×3×8^2=1/2×3×64=96m.
2. The motion of an object moving in a straight line is uniformly retarded so that it comes to rest in 20s after travelling a distance of 40m. What is the initial velocity of the body? SSCE.
Solution:
Data given; t=20s, s=40m, u=?, v=0 (body at rest.
S=t(v+u)/2
40=20(0+u)/2
80=0+20u (cross multiply).
U=80/20
U=4m/s.
MOTION OF A FREE FALL OR
MOTION UNDER GRAVITY
Acceleration 'a' is replaced by acceleration due to gravity 'g' when dealing with a body falling to the earth.
So anywhere you see 'a' in any of the equations of motion will change to 'g'. i.e
1st equation of motion; v=u+gt
3rd equation of motion: s=ut+1/2gt^2
4th equation of motion: v^2=u^2+2gs
Note:
* For all body moving upward, g is negative and v=0 i.e
1st equation of motion; v=u-gt
3rd equation of motion: s=ut-1/2gt^2
4th equation of motion: v^2=u^2-2gs
* If the body is released from height, u= 0 and s= h. From s=1/2at^2, therefore we have h=1/2gt^2..
* Provided that there is no air resistance, two bodies of different masses get to the ground at the same time. This obey the equation of time of flight which says:
t=√2h/g.
The equation reveals that their time of flight is independent on their masses.
Worked Example:
A palm fruit drops to the ground from the top of a palm tree 80m tall. How long does it take to reach the ground? (g=10m/s^2).
Solution:
Given that; h=80m, g=10m/s^2, t=?.
From h=1/2gt^2.
t=√2h/g=√2×80/10=√160/10=√16=4
t=4s.
Exercise:
Two grapefruit of masses 50g and 70g respectively are falling from the top of the tree simultaneously. Given that the height of the tree is 32m, calculate the time taken by the heavier fruit to hit the ground. Neglect air resistance. (g=10m/s^2)..
END OF LECTURE.
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Tell us if you find this helpful.
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