We have about three equation of linear motion introduced by sir
Isaac Newton’s in 1886, include the following
i)
Newton’s first
equation of motion
ii)
Newton’s second
equation of motion
iii)
Newton’s third
equation of motion
FIRST
EQUATION OF MOTION
It is related final velocity, initial velocity, acceleration and time
taken. The Newton’s first equation of motion is given by:
v = u + at
Consider a body with initial velocity (u), acceleration (a) for a
time (t) and final velocity (v)
SECOND
EQUATION OF MOTION
It relates distance (displacement) moved, initial velocity, time taken, and acceleration. The Newton
second equation is given by:
S = ut + ½(at2)
Consider the Distance = Average velocity x Time
THIRD
EQUATION OF MOTION
It relates to final velocity, initial velocity, acceleration and distance
moved. The Newton’s third equation of
motion is given by
v2 = u2 + 2as
Consider the first equation v = u + at
Hence the three equations of motion are
i)
v =
u + at
ii)
s =
ut + ½at2
iii)
v2
= u2 + 2as
Example 1
A body moving at 10 m/s accelerates uniformly
at 2 m/s2 for 8 seconds. Determine the final velocity.
Solution
Data given
Initial velocity (u) = 10 m/s
Acceleration (a) = 2 m/s2
Time (t) = 8 s
Final velocity (v) =?
Example 2
A body moving with a velocity of 30 m/s is
accelerated uniformly to a velocity of 50 m/s in 5 seconds.
Calculate;
a) Acceleration of the body
b) The distance travelled by the body
Solution
Data given
Initial velocity (u) = 30 m/s
Final velocity (v) = 50 m/s
Time (t) = 5 s
Acceleration (a) =?
Distance (s) =?
a)
Acceleration,
b)
The distance
travelled;
The distance travelled by the body =
200 m
Example 3
A body moving with a velocity of 5 m/s
accelerates at 2 m/s2 and covers a distance of 15 m. Determine the final
velocity just when covering the distance.
Solution
Data given
Initial velocity (u) = 5 m/s
Acceleration (a) = 2 m/s2
Distance (s) = 15 m
Final velocity (v) =?
The final velocity of the body = √85 m/s
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