### What is Average Speed ?

Average speed of a particle is the ratio of Total distance covered by a particle to the Total time taken .

$ \large Average \; Speed = \frac{Total \; distance \; covered}{Total \; time \; taken} $

A particle covers a distance by Δs = 10 m during a time interval Δt = 5 seconds.

Therefore $ \large \frac{\Delta s}{\Delta t}$ is the ratio of distance covered Δs and the time interval Δt . This ratio is termed as time rate of change of distance.

It is common known as Speed of the particle over that particular time interval Δt.

That is why this speed is known as “Average Speed” denoted mathematically as

$ u_{av} = \displaystyle \frac{\Delta s}{\Delta t}$

$ \displaystyle = \frac{10}{5} $

= 2 m/s

**Note : **If Speed is continuously changing with time then ,

$\displaystyle v_{avg} = \frac{\int v dt}{\int dt}$

This is called Time Average Velocity .

$\displaystyle v_{avg} = \frac{\int v dx}{\int dx}$

This is called Space Average Velocity .

Solved Example: When a body covers first half distance with speed v_{1} and next half distance with speed v_{2} then find the average speed ?

Solution: Let the total distance = S

Total time taken , t = t_{1} + t_{2 }

$ \displaystyle t = \frac{S}{2 v_1} + \frac{S}{2 v_2}$

$ \displaystyle Average \; Speed = \frac{Total \; Distance \;Travelled} {Total \;Time \;Taken}$

$ \displaystyle Average \; Speed = \frac{S}{ \frac{S}{2 v_1}+ \frac{S}{2 v_2} }$

$ \displaystyle Average \; Speed = \frac {2 v_1 v_2}{ v_1 + v_2 }$

### What is Instantaneous Speed ?

If the particle is not moving with uniform speed, the average speed completed in the above section depends on the time interval Δt and the instant when it was computed.

It becomes necessary to define a quantity, called **instantaneous speed** which does not depend on the time interval Δt .

We take the time interval Δt as it becomes very small (infinitesimal) and the corresponding distance travelled Δs (which is also infinitesimal).-

Δt → 0 (infinitesimal time)

Δs → 0 (infinitesimal distance)

As Δt → 0 , Δs → 0

#### Note : Instantaneous speed is equal to the magnitude of instantaneous velocity.

The ratio of the infinitesimal distance covered and the corresponding infinitesimal time yields a finite value. This finite value is known as the instantaneous speed of the particle.

### What is Uniform Speed ?

If the particle covers equal distances in equal intervals of time, it is said to be moving with uniform speed.

#### Student Notes:

# The slope of s-t graph at any time t gives the speed at that time t.

# The slope of x-t graph at any time t gives the instantaneous velocity (velocity at that time)

# For positive slope v is positive, for negative slope v is negative.

### What is Average Velocity ?

As the displacement of a particle changes with time, the ratio of change Δs^{→} of displacement vector s^{→} in a time interval Δt to that interval of time is known as time rate of change of displacement (vector) , this is known as the Velocity of the particle averaged over the time interval Δt or the average velocity,

$ \displaystyle \vec {v_(av) } = \frac{\vec\Delta S}{\Delta t} $

### What is Instantaneous Velocity ?

The velocity of an object at a particular instant or at a particular point of its path is called instantaneous velocity of the object.

$\large \vec{v} = \lim_{\Delta t \rightarrow 0} \frac{\vec{\Delta s }}{\Delta t}$

$ \displaystyle \vec{v} = \frac{\vec{ds}}{dt}$

* Instantaneous speed is the magnitude of instantaneous velocity

The slope of the v-t graph gives the acceleration. If the slope is positive the acceleration is positive; if the slope is negative then the acceleration is negative (retardation or deceleration).

the area of a-t graph gives the change in velocity. If the area remains (above x- axes) positive, the change in velocity is positive and vice versa.

### What is Uniform Velocity ?

A particle is said to be moving with uniform velocity if its velocity is independent of time.

Solved Example : A particle moves in a straight line obeying the relation x = t (t-1) where x = displacement in m and t = time in sec. Find the velocity of the particle when its displacement is zero.

Solution:

x = t (t − 1)

x = t^{2 }-t

If x = 0 , t ( t − 1) = 0

⇒ t = 0 and t = 1

The instantaneous velocity , $ \displaystyle v= \frac{dx}{dt}$ = $ \displaystyle \frac{d(t^2 -t)}{dt}$ = 2t – 1

Putting t = 0 and t = 1 we obtain

v_{t=0} = -1m/s , v_{t=1} = 1 m/s

Exercise : Referring to the previous example ,

(a) What is the significance of positive and negative values of instantaneous velocities ?

(b) What are the distance covered and displacement during t = 2 sec. ?

Solved Example : The displacement of a particle as a function of time is given by $\large x = t^3 – 2t^2 + 2t + 1$ , Find its velocity at t = 1s and at t = 2s.

Solution : $\large x = t^3 – 2t^2 + 2t + 1$

Differentiating with respect to time.

$\large \frac{dx}{dt} = 3 t^2 – 4t + 2 $

Velocity at t = 1 sec ,

$\large v= 3 \times 1^2 – 4 \times 1 + 2 = 1 m/s$

Velocity at t = 2 sec ,

$\large v= 3 \times 2^2 – 4 \times 2 + 2 = 6 m/s$

Solved Example : For a man who walks 720 m at a uniform speed of 2 *m/s*, then runs at a uniform speed of 4 *m/s* for 5 minute and then again walks at a speed of 1 *m/s* for 3 minutes. Find his average speed .

Solution:

S_{1} = 720 m , $\large t_1 = \frac{720}{2} = 360 sec$

S_{2} = 4 × 5 × 60 = 1200 m , t_{2} = 5 × 60 = 300 sec

S_{3} = 1 × 3 × 60 = 180 m , t_{3} = 180 sec

Average Speed $\large v_{avg} = \frac{S_1 + S_2 + S_3}{t_1 + t_2 + t_3}$

$\large v_{avg} = \frac{720 + 1200 + 180}{360 + 300 + 180}$

= 2.5 m/s