Cricket fan? Hockey fan? Soccer fan? What is the first thing that is taught when you first start training for these or any other sports? It is understanding the correct motion, speed acceleration or the Equations of Motion. Once you master the Equations of Motion you will be able to predict and understand every motion in the world.
Equations of Motion For Uniform Acceleration
As we have already discussed earlier, motion is the state of change in position of an object over time. It is described in terms of displacement, distance, velocity, acceleration, time and speed. Jogging, driving a car, and even simply taking a walk are all everyday examples of motion. The relations between these quantities are known as the equations of motion.
In case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore they can only be applied when acceleration is constant and motion is a straight line. The three equations are,
- v = u + at
- v² = u² + 2as
- s = ut + ½at²
where, s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time of motion. These equations are referred as SUVAT equations where SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (T)
Browse more Topics under Motion
- Introduction to Motion and its Parameters
- Graphical Representation of Motion
- Uniform Circular Motion
You can download Motion Cheat Sheet by clicking on the download button below
Derivation of the Equations of Motion
v = u + at
Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:
Acceleration = Change in velocity/Time Taken
Therefore, Acceleration = (Final Velocity-Initial Velocity) / Time Taken
Hence, a = v-u /t or at = v-u
Therefore, we have: v = u + at
v² = u² + 2as
We have, v = u + at. Hence, we can write t = (v-u)/a
Also, we know that, Distance = average velocity × Time
Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2
Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]
or s = (v² – u²)/2a
or 2as = v² – u²
or v² = u² + 2as
s = ut + ½at²
Let the distance be “s”. We know that
Distance = Average velocity × Time. Also, Average velocity = (u+v)/2
Therefore, Distance (s) = (u+v)/2 × t
Also, from v = u + at, we have:
s = (u+u+at)/2 × t = (2u+at)/2 × t
s = (2ut+at²)/2 = 2ut/2 + at²/2
or s = ut +½ at²
More Solved Examples For You
Example 1: A body starts from rest accelerate to a velocity of 20 m/s in a time of 10 s. Determine the acceleration of the boy.
Solution: Here, Final velocity v = 20 m/s and initial velocity u = 0 m/s (the body was at rest yo!). Therefore, Time taken t = 10 s. Hence, using the equation v = u +at.
a = (v-u )/t
= (20 – 0 ) /10
= 2 m/s2
Hence the acceleration of the body is 2 m/s2.
Example 2: A bus starts from rest and moves with constant acceleration 8ms−2. At. the same time, a car travelling with a constant velocity 16 m/s overtakes and passes the bus. After how much time and at what distance, the bus overtakes the car?
A) t =4s, s = 64m B) t = 5s, s = 72m C) t = 8s, s = 58m D) None of These
Solution: A) Let the position of the bus be PB and the position of the car be PC. From s = ut +½ at², we have
Since the initial velocity of the bus, u = 0, hence we have PB = ½ (8)t²
And PC = velocity × Time = 16×t. For the bus to overtake the car, we must have: PB = PC
Hence, ½ (8)t² = 16×t. Therefore, t = 4s.
Using the value of t = 4s in PB = ½ (8)t ², we have the position of the bus at the time of overtaking is = 64m.