## The Three Equations Of Linear Motion

- Posted by Max in Linear Motion |
- December 11th, 2009 |
- Comments

**Linear Motion** Let us begin our discussion on linear motion with few definitions.

**Velocity** Velocity of a moving body is the rate of change of position of the body in a straight line or the rate of displacement in a given direction. Velocity is expressed as meters per second (m/s). It is a vector quantity.

**Vectors:** Physical quantities that have both magnitude and direction are known as Vectors.

Vectors are represented by a straight line with an arrow. Examples of vectors are Displacement, Velocity, Acceleration and Force.

Vector quantities are represented by an arrow over them.

**A ————1————–1————→ B **East

The above figure represents a velocity of 3 meters per second. The magnitude is 3 units of length per second in the direction AB, i.e., East.

**Scalars:** Quantities that have a magnitude but no direction are known as Scalars. Examples of scalars are Speed, Mass, Volume, Density etc. Suppose a person starts from A and moves along a curved path ACB of distance S meters and reaches B in‘t’ seconds, then the average speed of the person is S/t m/s.

However, the displacement is only AB, which is the straight line distance from A to B that has taken place in ‘t’ seconds. Hence, the velocity of the person is only equal to AB/t m/s, that is very much different from the Speed S/t .

Uniform and variable velocity: Velocity of a moving body is said to be uniform if the body moves equal distances in equal interval of time or if velocity is same at all points of the motion.

On the other hand, if the velocity changes with time, then it is variable velocity.

**Acceleration** **Acceleration is defined as the rate of change of velocity, which means how much the velocity changes per second. Acceleration is said to be uniform if the change of velocity is same throughout. If it is not, then acceleration is variable.**

Acceleration is measured in meter per second per second (m/s/s) or in a shorter form m/s^{2}. If the velocity increases every second, then the acceleration is positive. If it decreases every second, then it is negative acceleration or retardation.

Motion in a straight line is known as Linear Motion

**Equation 1: This equation provides the relation between initial velocity, final velocity, the time taken to reach it, and the acceleration. **

Let the initial velocity be u, the final velocity v, time taken t, and acceleration ‘a’, then the change of velocity = v – u

The time taken for this change = t

Therefore, the rate of change of velocity, i.e., change of velocity per sec

or acceleration a = (v – u)/t i.e., **v = u + at **

Note: If there is retardation, we use ‘–a’ in place of ‘a’.

**Equation 2: This equation provides the relation between distance travelled, initial velocity, acceleration and time.**

Let the initial velocity be u

Let the final velocity be v

Let the time taken be t

And, let the distance travelled be S

Now S = average velocity × time = [(u + v)/2]/t

But v = u + at {from equation 1}

Substituting for v, we get S = [(u + u + at)/2]×t

= ut + at^{2}/2

Hence, **S = ut + at ^{2}/2**

The same formula can be used for retardation also, but we need to put ‘-a’ in place of ‘a’.

**Equation 3: This equation provides the relation between initial velocity, final velocity, acceleration and distance. There is no time factor in it.**

v = u + at

Squaring, we get v^{2} = (u + at) ^{2}

= u^{2} + 2uat +a^{2} t^{2}

= u^{2} + 2a (ut + a t^{2}/2)

= u^{2} + 2aS

Hence, **v ^{2} = u^{2} + 2aS** In case of retardation, ‘a’ becomes ‘-a’

Note: To find the distance travelled in a given time, we make use the formula

S = ut + at^{2}/2.

If it is required to find the distance travelled in the third second, then using the formula, we find the distance travelled in 3 sec and in 2 sec and then subtract the value of the second from the first.

However, we can find a single formula to find the distance travelled in the‘t’ th second as given below:

**To find the distance travelled in the‘t’ th second**

We have to first find the distance in t sec and in (t-1) sec, subtract and simplify. We will get the required formula.

S_{t} = ut + at^{2}/2, where S_{t} is the distance travelled in t sec.

S_{t-1} = u(t-1) + a(t-1)^{2}/2 where S_{t-1} is the distance travelled in t-1 sec.

S_{t} – S_{t-1} = ut + at^{2}/2 – u(t-1) – a(t-1)^{2}/2

i.e., S_{’t’th}= ut –u(t-1) + a[t^{2} - (t-1)^{2}]/2

= u + a(t^{2} – t^{2}+ 2t – 1)/2

= u + a(2t – 1)/2

= u + a(t – 1/2)

By using the above formula, we can calculate the distance traveled in any particular second.

Thus, when u = 2m/s and a = 0.5 m/s^{2}, the distance travelled in the 3rd sec is S_{3rd} = 2 + 0.5 (3 – 1/2) = 3.25 m

**All freely falling bodies are subjected to a uniform acceleration. Its values at the equator and the poles are different. Its value is usually taken to be 9.8 m / s ^{2} which is denoted by ‘g’. The three equations of linear motion apply to falling bodies as well. To apply, we have to replace ‘g’ for ‘a’.
Thus the three equations of motion in case of falling bodies are **

**V = u + gt**

S = ut + gt^{2}/2

v^{2} = u^{2} + 2gS

**If a body is projected upwards, we have to use ‘-g’ for ‘g’. **

Example: A body starting from rest and executing an accelerated motion covers a distance of 9 cm in 6 seconds. Calculate (a) the acceleration (b) the final velocity.

(a) S = ut + at^{2}/2

Implies, 9 = (0×6) + (a×6^{2})/2

Therefore, a = 0.5 m/s^{2}

(b) V = u + at = 0 + (0.5×6) = 3 m/s

A bullet is fired vertically with an initial velocity of 29.4 m/s (a) How high will it reach? (b

) What is the time taken to reach that height?

u = 29.4 m/s; v = 0

g = -9.8 m/s^{2}

S = ?

t = ?

(a) v^{2} = u^{2} + 2gS

Implies, 0 = 29.4^{2} + 2×(-9.8)×S

Implies, S = 2×9.8 = (29.4)^{2}

Implies, S = (29.4)^{2}/(2×9.8) = 44.1 m

(b) v = u + gt

0 = 29.4 + (-9.8) t

Therefore, t = 3 sec

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