Grade 11 Physics Note


Speed and velocity:


The time rate of change of displacement is known as velocity. It is a vector quantity and the magnitude of velocity is equal to that of displacement.  The direction of velocity is along the direction of displacement.  Suppose if d is the total displacement travelled by the body in time t, then its average velocity during time t will be:

${{\rm{v}}_{{\rm{avg}}}} = \frac{{\rm{d}}}{{\rm{t}}}$


The rate at which the object covers a distance is called as its speed.  It is a scalar quantity.  For example a fast moving object will have a high speed and it will cover more distance in less time whereas, a slow moving object will have less speed and it will cover less distance in that time

   ${\rm{averagespeed}} = \frac{{{\rm{distance\: travelled}}}}{{{\rm{timetaken}}}}$


Rest and motion:


If an object changes its position with respect to it’s surrounding in a specific time.  Then the body is said to be in a state of motion.

                                              ${\rm{motion}} = \frac{{\Delta {\rm{d}}}}{{{\rm{time}}}}$


If an object does not change its position with respect to its surrounding then the body is said to be in the state of rest.

Uniform and non – uniform acceleration:

Uniform acceleration:

If the velocity v of a body changed by equal amount in equal intervals of time t the body is said to have uniform acceleration for a body moving with uniform acceleration its average acceleration is equal to the instantaneous acceleration.

Non-uniform acceleration:

If the velocity v of the body changes in unequal amount in unequal intervals f time then the body is said to have non -uniform acceleration.


Equations derivation:

(i) S = ut + $\frac{1}{2}$at2

 (ii) v2 = u2 + 2as.

(i) Let ‘S’ be the distance, ‘u’ be the initial velocity, ‘v’ be the final velocity, ‘a’ be the acceleration and ‘t’ be the time. To derive the equation of motion we have the velocity which is expressed as,

${\rm{\bar v}}$= $\frac{{\Delta {\rm{x}}}}{{\Delta {\rm{t}}}}$

Where Δx refers to the change in displacement in a time interval Δt

Rearranging for displacement,

Δx = $\overline {{\rm{v\: }}} $Δt

x – xi = ${\rm{\bar v}}$Δt

x = xi + ${\rm{\bar v}}$Δt – equation 1

The average velocity is given as the mean of initial and final velocities,

${\rm{\bar v}}$ = $\frac{1}{2}$ (u + v) – equation 2

From first equation of motion we have,

v = u + aΔt

Substituting the value of ‘v’ in equation 2,

${\rm{\bar v}}$ = $\frac{1}{2}$ (u + v)

${\rm{\bar v}}$ = $\frac{1}{2}$ (u + u + a Δt)

${\rm{\bar v}}$ = $\frac{1}{2}$ (2u + a Δt)

${\rm{\bar v}}$ = u + $\frac{1}{2}$ a Δt – equation 3

Now substituting equation 3 in equation 1 we get,

x = xi + ${\rm{\bar v}}$Δt

x = xi + (u + $\frac{1}{2}$ a Δt) Δt

x = xi + u Δt + $\frac{1}{2}$ a Δt2

xi is the initial displacement and may be neglected when we do not consider initial displacement and we can represent displacement by ‘s’. Further Δt refers to a time span and can be written simply as t. So this equation of motion can be written as,

s = u t + $\frac{1}{2}$ a t2


(ii) This equation of motion can be derived using 1st and 2nd equations of motion. We have 1st equation of motion as,

v = u + a t

Rearranging for time we get,

t = $\frac{{{\rm{v\: }} - {\rm{\: u\: }}}}{{\rm{a}}}$ – equation 1

And we have 2nd equation of motion as,

s = u t + $\frac{1}{2}$ a t2 – equation 2

Substituting value of t from equation 1 into equation 2 we get,

s = u $\frac{{{\rm{v\: }} - {\rm{\: u\: }}}}{{\rm{a}}}$ + $\frac{1}{2}$a $(\frac{{{\rm{v\: }} - {\rm{\: u\: }}}}{{\rm{a}}}$)2

a s = u v – u2 + $\frac{1}{2}$ (v2 – 2 u v + u2)

a s = $\frac{1}{2}$ (v2 – u2)

This finally becomes,

2 a s = v2 – u2


Relative velocity:

Relative velocity is defined as the velocity of one of the object with respect to the another object.

 If the two bodies A and B is travelling with velocity ${{\rm{V}}_{\rm{A}}}$ and ${{\rm{V}}_{\rm{B}}}{\rm{\: \: in\: same\: \: direction\: }}$ then the resultant velocity or the velocity of the A w.r.t B is given by   ${{\rm{V}}_{{\rm{AB}}}}{\rm{\: }} = {{\rm{V}}_{\rm{A}}} - {\rm{\: }}{{\rm{V}}_{\rm{B}}}$

If the two bodies A and B is travelling with velocity ${{\rm{V}}_{\rm{A}}}$ and ${{\rm{V}}_{{\rm{B\: \: }}}}{\rm{in\: opposite\: direction\: \: }}$ then the resultant velocity or the velocity of the A w.r.t B is given by   ${{\rm{V}}_{{\rm{AB}}}}{\rm{\: }} = {{\rm{V}}_{\rm{A}}} - {\rm{\: }}\left( { - {{\rm{V}}_{\rm{B}}}} \right)$  =  ${{\rm{V}}_{\rm{A}}} + {\rm{\: }}{{\rm{V}}_{\rm{B}}}$

When the two bodies are making an acute angle to obtain the relative velocity of an object with respect to the observer parallel law of vectors and triangle law of vectors can be used to calculate the relative velocity.


Horizontal range of the projectile and its path is parabola:

Let us consider the object which is projected with initial velocity u with an angle α with ground (x-axis). The velocity has two components, ucos α along the x-axis and using α along Y – axis.

The motion of projectile is two dimensional. The horizontal distanced covered is not affected by the gravity whereas the vertical distance is affected by the gravity.

Let us consider the objects reaches at point P in time t whose horizontal and vertical distance are x and y and it is given by  ${\rm{x}} = {\rm{ucos\: }}\alpha {\rm{\: \: *t}}$ ,

${\rm{t}} = \frac{{\rm{x}}}{{{\rm{ucos\: }}\alpha {\rm{\: \: }}}}$  ……………1

${\rm{y}} = {\rm{\: \: usin\: }}\alpha {\rm{\: \: *t}} - \frac{1}{2}{\rm{\: g}}{{\rm{t}}^2}{\rm{\: }}$

${\rm{y}} = {\rm{\: \: usin\: }}\alpha {\rm{\: \: *}}\frac{{\rm{x}}}{{{\rm{ucos\: }}\alpha {\rm{\: \: }}}}{\rm{\: \: }} - \frac{1}{2}{\rm{\: g}}{\left( {\frac{{\rm{x}}}{{{\rm{ucos\: }}\alpha {\rm{\: \: }}}}{\rm{\: \: }}} \right)^2}$

${\rm{y}} = {\rm{\: \: usin\: }}\alpha {\rm{\: \: *}}\frac{{\rm{x}}}{{{\rm{ucos\: }}\alpha {\rm{\: \: }}}}{\rm{\: \: }} - \frac{1}{2}{\rm{\: g}}{\left( {\frac{{{{\rm{x}}^2}}}{{{{\rm{u}}^2}{{\cos }^2}{\rm{\: }}\alpha {\rm{\: \: }}}}{\rm{\: \: }}} \right)^{}}$

This is the equation of parabola

This shows that the path followed by the object is parabolic.

For the horizontal range R ,

We have to calculate the time by which projectile remains in the air

We have S $ = {\rm{ut\: }} + \frac{1}{2}{\rm{g}}{{\rm{t}}^2}$

 0 = usin α    + ½ (-g) T2  (since vertical displacement is 0 )

T = $\frac{{2{\rm{usin\: }}\alpha {\rm{\: \: }}}}{{\rm{g}}}$

Horizontal range = horizontal velocity * time of flight (T)

 R = ucos α  * 2usin α  /g


The path of projectile projected horizontally from the top of a tower is parabolic:

When an object is thrown into the atmosphere, it falls under the effect of gravity alone .This is called projectile.

Let us consider an object projected horizontally from a height h from the ground with initial velocity u.

Let the position of an object at point p after time t is given by

For horizontal distance

X = ut

t = $\frac{{\rm{x}}}{{\rm{u}}}$

For vertical distance

y = 0 + $\frac{1}{2}{\rm{g}}{{\rm{t}}^2}$

Or, y =  $\frac{1}{2}{\rm{g\: }}{\left( {\frac{{\rm{x}}}{{\rm{u}}}} \right)^2}{\rm{\: }}$

This is the expression for the path of horizontal projectile and its shows that it is parabolic.

Although particle has initial velocity it gains  vertical velocity  due to the gravity .If ${{\rm{V}}_{\rm{x}}}{\rm{\: \: and\: }}$${{\rm{V}}_{\rm{y}}}$  are the componenet of the velocity at  P in horizontal and vertical direction  the resultant velocity is given by  V = $\sqrt {{\rm{V}}_{\rm{x}}^2 + {\rm{\: V}}_{\rm{y}}^2} $

Since Vx is the consant being no acceleration in horizontal direction

                                                                      ${{\rm{V}}_{\rm{y}}} = {\rm{gt}}$

V = $\sqrt {{\rm{V}}_{\rm{x}}^2{\rm{\: \: }} + {{\left( {{\rm{gt}}} \right)}^2}} $

If α is the angle made by the resultant velocity V with the horizontal, then

Tanα =  $\frac{{{{\rm{v}}_{\rm{y}}}}}{{{{\rm{V}}_{\rm{x}}}}}$  = $\frac{{{\rm{gt}}}}{{\rm{u}}}$

$\alpha  = {\tan ^{ - 1}}\frac{{{\rm{gt}}}}{{\rm{u}}}{\rm{\: \: }}$


A method for calculation of velocity of a body A relative to another body B when they are moving making an angle θ:

Let ${{\rm{v}}_1}{\rm{\: and\: }}{{\rm{v}}_2}{\rm{\: }}$a re the velocity of body A and body B and the resultant velocity ${{\rm{v}}_{{\rm{AB}}}}$

Here,${\rm{\: }}\mathop\to \limits_{{{\rm{V}}_1}} $    and ${\rm{\: \: }}\mathop\to \limits_{{{\rm{V}}_2}} $    are represented by the sides of parallelogram OBCA and the${\rm{\: \: }}\mathop\to \limits_{\rm{V}} $     is represented by OC such that   ${\rm{\: \: }}\mathop\to \limits_{\rm{v}} $    = $\mathop\to \limits_{{{\rm{V}}_1}} $    +${\rm{\: }}\mathop\to \limits_{{{\rm{V}}_2}} $   .let the angle between $\mathop\to \limits_{{{\rm{V}}_1}} {\rm{\: \: }}$and $\mathop\to \limits_{{{\rm{V}}_2}} $    is ${\rm{\: \: }}\theta $

Magnitude of  $\mathop\to \limits_{\rm{v}} $

To calculate $\mathop\to \limits_{\rm{v}} $    , draw the perpendicular from point C to D as shown in figure.

In $\Delta $OCD

${\rm{O}}{{\rm{C}}^2} = {\left( {{\rm{OA}} + {\rm{AD\: }}} \right)^2} + {\rm{C}}{{\rm{D}}^2}$                ………………………i


CD = ${{\rm{V}}_2}{\rm{\: }}$ sinθ

AD= ${{\rm{V}}_2}$cosθ

 From i

${\rm{O}}{{\rm{C}}^2}{\rm{\: }} = {\rm{\: \: }}{\left( {{{\rm{v}}_1} + {{\rm{v}}_2}{\rm{cos}}\theta {\rm{\: }}} \right)^2} + {\left( {{{\rm{v}}_2}{\rm{sin}}\theta {\rm{\: }}} \right)^2}$

${{\rm{V}}_{{\rm{AB}}}}{\rm{\: }} = {\rm{\: }}\sqrt {{\rm{\: }}{{\rm{v}}_1}{{\rm{\: }}^2} + {\rm{\: }}2{{\rm{v}}_1}{{\rm{v}}_2}{\rm{cos}}\theta {\rm{\: }} + {\rm{\: \: }}{{\rm{v}}_2}{{\rm{\: }}^2}} $

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