**Category : **9th Class

The motion of the body moving along a straight line with uniform acceleration can be described by three equations of motion. These equations can be derived as follows.

**First Equation of Motion**

Let us consider an object moving with a initial velocity 'u'. If it is subjected to a uniform accleration 'a' such that it attains a velocity of V after time 't', then

Acceleration \[=\frac{Final\,velocity\,-\,initial\,ve\operatorname{lo}city}{Time\,taken}\]

So, \[a=\frac{v-u}{t}\]

\[\Rightarrow \,\,at=v-u,\] or \[v=u+at\]

where, v = final velocity of the body

u = Initial velocity of the body

a = acceleration

and, t= time taken

**Second Equation of Motion**

The second equation of motion is: \[s=ut+\frac{1}{2}\,a{{t}^{2}}\]. It gives the distance traveled by a body in time t.

Let us derive this second equation of motion:

Let us consider an object moving with a initial velocity **'u'** and a uniform acceleration **"a"**. Let it attains a velocity V after some time **'t'**. Let the distance traveled by the object in this time be **'s'**. The distance traveled by a moving body in time **'t'** can be found out by considering its average velocity. Since the initial velocity of the body is **'u'** and its final velocity is V, the average velocity is given by:

**Average velocity \[=\frac{\text{Initial velocity }+\text{ final velocity}}{2}\] **

i.e., Average velocity \[=\frac{u+v}{2}\]

Also,** Distance traveled = Average velocity \[\times \] Time**

so, \[S=\,\left( \frac{u+v}{2} \right)\times t\] …(1)

From the first equation of motion we have, **v = u + at**. Putting this value of **‘v’** in equation (1), we get:

Or \[S=\,\left( \frac{u+v}{2} \right)\times t\]

Or \[S=\,\left( \frac{2u+at}{2} \right)\times t\]

Or \[S=\frac{2ut+a{{t}^{2}}}{2}\]

Where s = distance traveled by the body in time t

u = Initial velocity of the body and, a = Acceleration

**Third Equation of Motion**

The third equation of motion is: v- u^ 2as. It gives the velocity acquired by a body in traveling a distance 's'.

The third equation of motion can be obtained by eliminating -f from the first two equations of motion and using the second equation of motion

From the second equation of motion we have:

\[s=ut+\frac{1}{2}\,a{{t}^{2}}\] …(1)

And from the first equation of motion we have :

\[v=u+at\]

Or, \[at=v-u\]

Or, \[t=\frac{v-u}{a}\]

Putting this value of t in equation (1), we get:

\[S=u\,\left( \frac{v-u}{a} \right)+\frac{1}{2}a\,{{\left( \frac{v-u}{a} \right)}^{2}}\]

Or \[S=u\,\left( \frac{v-u}{a} \right)+\frac{1}{2}\,\frac{{{(v-u)}^{2}}}{a}\]

Or \[S=\frac{uv-{{u}^{2}}+{{v}^{2}}+{{u}^{2}}-2uv}{2a}\]

Or \[2as={{v}^{2}}-{{u}^{2}}\]

Or \[{{v}^{2}}={{u}^{2}}+\,2as\]

Where, v = final velocity,

u = initial velocity,

a = acceleration

and s = distance traveled

**Graphical Method of Finding Equations of Motion**

We can derive the equation of motions using the velocity time graph. Consider the motion of the object moving along a straight line with a uniform acceleration. The above graph shows the motion of an object along a straight line with a uniform accleration **'a'**. The point A and B on the graph corresponds to time 0 and t respectively.

Let **u** be the velocity of the object at time t = 0 represented by line OA on the graph. Also let the velocity of the particle be vat time t represented by line OD on the graph.

**First Equation of Motion**

The slope of the velocity-time graph gives the acceleration of an object moving along a straight line.

For line AB, the slope is given by

Slope \[=\frac{BC}{AC}\]

\[\Rightarrow \,\,a=\frac{BC}{AC}=\frac{BE-CF}{OE}\]

\[\Rightarrow \,\,a=\frac{v-u}{t-0}=\frac{v-u}{t}\]

\[\Rightarrow \,\,v=u+at\]

\[\mathbf{=v=u+at}\]

**Second equation of Motion**

The area under the velocity-time graph is equal to the displacement. In the time interval 0 - t, displacement = area OABE

S = area OABE = area of the rectangle OACE + area of the triangle ABC

\[\Rightarrow \] \[S=OA\times OE+\frac{1}{2}\times AC\times BC\]

\[\Rightarrow \] \[S=OA\times OE+\frac{1}{2}\times AC\times (BC-CE)\]

\[\Rightarrow \] \[S=u\times t+\frac{1}{2}\times t\times (v-u)\] (using first equation of motion)

\[\Rightarrow \] \[S=u\times t+\frac{1}{2}\times a\times {{t}^{2}}\]

or \[\Rightarrow \,\,S=ut+\frac{1}{2}\,a{{t}^{2}}\]

**Third equation of motion**

The area under the \[v-t\] graph is =area of trapezium AOEB Area of trapezium \[=\frac{1}{2}\] (sum of parallel sides) \[\times \] (altitude)

Or, Area = \[=\frac{1}{2}\times \,(OA+BE)\times AC\]

Or, \[S=\frac{1}{2}\times (u+v)\times t\]

Or, \[S=\frac{1}{2}\times (u+v)\times \,\left( \frac{v-u}{a} \right)\]

Or, \[{{v}^{2}}={{u}^{2}}+2as\]

**Average, velocity for uniformly Accelerated Motion**

The average velocity of an object can be derived using the first and second equation of the motion. Let us consider an object moving along a straight line with a uniform accleration 'a'. The average velocity can be defined as,

Average velocity \[=\frac{\text{Total Displacement}}{\text{Total Time}}\]

Or, \[\frac{S}{t}=\frac{{{v}^{2}}-{{u}^{2}}}{2at}\]

\[{{V}_{av}}=\frac{v+u}{2},\] (using first equation of motion)

**An object moving with a velocity of 15 m/s decelerate at the rate of 1.5 m/ s ^{2} Find the time taken by the objects to come to rest.**

(a) 10 sec

(b) 9.5 sec

(c) 11 sec

(d) 12.2 sec

(e) None of these

** **

**Answer: (c)**

**Explanation**

We have, from first equation of motion

\[t=\frac{v-u}{a}\]

Or, \[t=\frac{0-15}{-1.5}=\frac{15}{1.5}=10\,\,\sec \]

**A car accelerates from 15 km/h to 60 km/h in 300 seconds. Find the distance traveled by the car during this time.**

(a) 3.35km

(b) 3.33km

(c) 4.33km

(d) 5km

(e) None of these

** **

**Answer: (d)**

**Explanation**

We have, u = 15 km/h, v = 60 km/h and \[t=300\] \[\sec =\frac{1}{12}\]

From first equation of motion,

\[v=u+at,\]

At \[t=0,\] the velocity is u = 20 km/h

At t = 4 min \[=\frac{1}{15}h,\] the velocity is v = 80 km/h

Using \[v=u+at,\]

\[a=\frac{v-u}{t}=a=\frac{60-15}{\frac{1}{12}}\]

Now the distance covered by the car is given by,

\[S=ut+\frac{1}{2}a{{t}^{2}}\]

\[S=15\times \frac{1}{12}+\frac{1}{2}\times 540\times \frac{1}{12}\times \frac{1}{12}=5\,\text{km}\]

**Circular Motion **

Where, T is the time taken by the object to move around the circular path.

- The motion of the blades of an electric fan around the axle.
- The motion of an electron around the nuclear of an atom.
- The motion of a satellite around the earth.
- A stone tied at one end of the string and whirled above the head of a person in circular path.

**Uniform Circular Motion**

When a body moves along a circular path, with constant speed then its direction of motion (or direction of speed) keeps changing continuously.

Circle can be considered to be polygon with infinite sides.

Since the velocity changes due to continuous change in direction, therefore the motion along a circular path is said to be accelerated. When a body moves in a circular path with uniform speed (constant speed), its motion is called** uniform circular motion**..

- The tip of a seconds' hand of a watch exhibits uniform circular motion on the circular dial of the watch. Please note that though the speed of the tip of seconds' hand is constant, but its velocity is not constant, because the direction of motion of tip of seconds' hand changes continuously. Thus, the motion of the tip of seconds' hand of a watch is accelerated.
- An anthele (or cyclist) moving on a circular track with a constant speed exhibits uniform circular motion. This motion is accelerated because of a continuous changes.

**An artificial satellite is moving in a circular orbit of radius 32,000 km. If it takes 30 hours to complete one revolution around the earth, then find the velocity of the satellite.**

(a) 9428.57 km/h

(b) 6704.76 km/h

(c) 9500 km/h

(d) 7256.68 km/h

(e) None of these

** **

**Answer: (b)**

**Explanation **

We have, R =32,000 km and t = 30 hours

Circumference of the orbit \[=2\pi \,\,r\,b=\,2\times \frac{22}{7}\times 32,000=201,142.86\,\text{km}\]

Now, velocity \[=\frac{2\pi R}{T}=\frac{201142.86}{30}=\,6704.76\,km/h\]

**A cyclist takes 180 second to complete one round of the circular track. A the radius of the circular track is 90 metres, then calculate his speed. (Given \[\pi =\frac{22}{7}\])**

(a) 3.14 m/sec

(b) 3.56 m/sec

(c) 4.25 m/sec

(d) 5.14 m/sec

(e) None of these

** **

**Answer: (a)**

**Explanation **

We have \[t=180\,\,\sec \] and \[r=90\] metres

\[V=\frac{2\pi r}{t}\]

Or, \[V=\frac{2\times \frac{22}{7}\times 90}{180}=3.14\,\,m/\sec \]

- Continous change of position of an object is called
**motion**. - The actual length of path covered by an object is called
**distance**. - The rate of change of velocity is called
**acceleration**. - If an object covers equal distance in equal time interval, it is called
**uniform motion**. - The distance time graph for uniform motion is a straight line.
- The motion of an object on a circular path is called
**circular****motion**.

*play_arrow*Motion*play_arrow*Distance and Displacement*play_arrow*Distance -Time Graph for Uniform Motion*play_arrow*Acceleration*play_arrow*Equations of Motion

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