- distance and displacement
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- Position
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- Velocity and speed
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- Instantaneous velocity and speed
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- Acceleration
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- Kinematic equations for uniformly accelerated Motion
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- Free fall acceleration
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- Relative velocity
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- Solved Examples Part 1
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- Solved Examples Part 2
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- Solved Examples Part 3
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- Solved Examples Part 4
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- Solved Examples Part 5

Velocity is defined as the rate of change of displacement.

- It is a vector quantity, both magnitude and direction are required to define it.
- It's direction is same as that of displacement.
- SI unit of velocity is m/s

Speed is defined as the rate of change in distance with respect to time.

- It is a scalar quantity. Only magnitude is required to define the speed.
- Speed and velocity both have same unit.

Now sometimes students ask questions like

Average Speed = Total distance/Total time

Average Velocity = Total displacement/Total time

Just like both displacement and distance are different, average speed and average velocity are also different for ex. Motion around a circular track.

- Consider a particle undergoing motion along a straight line i.e. particle is moving along X-axis.

- Here in this case X co-ordinate describing motion of the particle from origin $O$ varies
with time or we can say that X co-ordinate depends on time.

- If at time t=t
_{1}particle is at point P , at a distance x_{1}from origin and at time t=t_{2}it is at point Q at a distance x_{2}from the origin then displacement during this time is a vector from point P to Q and is

$\Delta x = {x_2} - {x_1}$ (1)

- The
**average velocity**of the particle is defined as the ratio of the displacement Δx of the particle in the time interval Δt=t_{2}-t_{1}. If v_{avg}represents average velocity then,

- Figure (5b) represents the co-ordinate time graph of the motion of the particle i.e., it shows how the value of x-coordinate of moving particle changes with the passage of time.
- In figure (5b) average velocity of the particle is represented by the slope of chord PQ which is equal to the ratio of the displacement Δx occurring in the particular time interval Δt.
- Like displacement average velocity
**v**also has magnitude as well as direction i.e., average velocity is a vector quantity._{avg} - Average velocity of the particle can be
*positive as well as negative*and its positive and negative value depends on the sign of displacement. - If displacement of particle is zero its average velocity is also zero.
- Graphs below shows the x-t graphs of particle moving with positive, negative average velocity and the particle at rest.

Average velocity is defined as the ratio of total displacement to total time. It is the average values of the velocities we know.

This is extra section where we talk about **average velocity formula**. From average velocity definition we know that it is the ratio
between total displacement and total time but this $v_{av}$ i.e., average velocity formula can have different form according to different problems.
So before solving problems that asks you to find average velocity please check under which case your problem falls.

**Case 1:-**

If $X$ is total displacement of the body and $t$ is the total time taken by the body to complete the displacement.The average velocity $V_{av}$ is given by

$$v_{av} = \frac{Total\ distance\ traveled}{total\ time\ taken} = \frac{X}{T}$$**Case 2:-**

If we know that body have initial displacement position $x_0$ at time $t_0$ undergoes a motion and attains final displacement $x_f$ at time $t_f$ then, the average velocity is given as

$$v_{av}=\frac{x_f - x_0}{t_f - t_0}=\frac{\Delta\ x}{\Delta\ t}$$**Case 3:-**

If we have a knowledge that $x_1, x_2, . . . . . x_n$ are the distance covered by the body for time $t_1, t_2, . . . . . t_n$. The average velocity is

$$ \frac{x_1 + x_2 + . . . . . . . + x_n}{t_1 + t_2 + . .... . . + t_n}$$-
**Case 3:-**

If a velocity is changing as a result of a constant acceleration, the average velocity can be found by adding the initial and final velocities, and dividing by 2. Here it is important to note that this formula applies only for constant acceleration case.

Consider a body moving with uniform acceleration along a straight path. At some instant of time, initially its velocity is $U$ m/s and at the end of t seconds let its final velocity be $V$ m/s. Hence the average velocity is given as

$$ v_{av}=\frac{v+u}{2}$$**How to derive this average velocity formula**

Average velocity is displacement divided by time. Let us start with the formula for displacement and divide it by t.

\begin{align} s=ut+\frac{1}{2}at^{2} \\ average \quad velocity & = \frac{ut+\frac{1}{2}at^{2}}{t} \\ & =u+\frac{1}{2}at \\ & =\frac{2u+at}{2} \\ & =\frac{u+(u+at)}{2} \end{align}

now, $v=u+at$

$$ average \quad velocity = \frac{u+v}{2} $$

- From graph 5c it is clear that for positive average velocity slope of line slants upwards right or we can say that it
has
*positive slope*. - For negative average velocity slope line slants upwards down to the right i.e. it has
*negative slope*. - For particles at rest slope is
*zero*. - So far we have learned that Average speed is defined as total distance traveled divided by time taken.
- Displacement of the object is different from the actual distance traveled by the particle.
- For actual distance traveled by the particle its average speed is defined as the total distance traveled by the particle in the time interval during which the motion takes place.
- Mathematically,

- Since distance traveled by an particle does not involve direction so speed of the particle depending on distance traveled does not involve direction and hence is a scalar quantity and is always positive.
- Magnitude of average speed may differ from average velocity because motion in case of average speed involve
distance which may be greater than magnitude of displacement.

Let us consider an example

here a man starts traveling from origin till point Q and return to point P then in this case displacement of man is

Displacement from O to Q is OQ=80m

Displacement from Q to P is =20m-80m=-60m

total displacement of particle in moving from O to Q and then moving Q to P is = 80m + (-60m) = 20 m

Now total distance traveled by man is $OQ+OP= 80m +60m = 140m$

Hence during same course of motion distance traveled is greater then displacement. - From this we can say that average speed depending on distance is in general greater than magnitude of velocity.

Let us now consider some of the solved examples for the concepts Average speed and Average velocity. Questions are important for understanding the concept. Try to understand these solved examples and then try to solve few questions on your own.

Time taken to travel first half distance ${t_1} = \frac{{x/2}}{{40}} = \frac{x}{{80}}hr$

Time taken to travel rest half distance ${t_2} = \frac{{x/2}}{{60}} = \frac{x}{{120}}hr$

Therefore Average speed = (Total distance)/(Total time) = $\frac{x}{{(x/80) + (x/120)}} = 48Km/hr$

$\frac{s}{3} = {v_0}{t_1}$

${t_1} = \frac{s}{{3{v_0}}}$ (1)

and,

$\frac{{2s}}{3} = {v_1}\left( {\frac{{{t_2}}}{2}} \right) + {v_2}\left( {\frac{{{t_2}}}{2}} \right)$

or,

${t_2} = \frac{{4s}}{{3({v_1} + {v_2})}}$ (2)

Average Velocity $ = \frac{s}{{{t_2} + {t_1}}}$

or, Average Velocity $ = \frac{s}{{\left( {\frac{s}{{3{v_0}}}} \right) + \left( {\frac{{4s}}{{3({v_1} + {v_2})}}} \right)}} = \frac{{3{v_0}({v_1} + {v_2})}}{{{v_1} + {v_2} + 4{v_0}}}$

Average speed is defined as total distance traveled divided by time taken and Average velocity is defined as change of displacement divided by the time taken. Now since distance traveled by any particle is either grater then the displacement or it is equal to the displacement , the velocity would be given as

${v_{av}} \geq |{{\vec v}_{av}}|$

that is usually average speed is greater than the magnitude of average velocity.

For example if a body returns to its starting point after some motion, then as distance travel would be finite while displacement would be zero. So in this case average speed would be greater then zero but magnitude of average velocity would be equal to zero.

However in case of motion along the straight line without change in direction, magnitude of displacement would be equal to distance and two definitions would mean the same.

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