Question

Derive all the three equations of motion in graphical method
and also in alternate method

Answer 1

Answer:

Graphical method  :  Consider an object moving along a straight line with uniform acceleration a let u be the initial velocity of the object at time t=0 and v be the final velocity of the object at time t . let s be the distance travelled by the object in time t.

velocity -time graph of this motion is a straight line PQ as shown in the figure  

where OP=u=RS

OW=SQ=v and  

OS=PR=t

(I) for the first equation of motion :  we know that the slope of velocity -time graph of uniformly accelerated motion represents the acceleration of the object  

i.e. acceleration = slope of the velocity -time graph PQ

or a=  

PR

QR

​  

=  

OS

QR

​  

 

a=  

OS

SQ−SR

​  

=  

t

v−u

​  

 

or v−u=at  

or v=u+at .....(i)  

This is the first equation of uniform acceterated motion.

(ii) Second equation of motion ;  We know that the area under the velocity-time graph for a given time interval represents the distance covered by the uniformly accelerated object in that interval of time.

∴ Distance (displacement) travelled by the object in time t is :  

S = area of trapezium OSQP

= area of rectangle  OSRP + Area of triangle PRQ

or S=OS×OP+  

2

1

​  

×PR×PQ

(area of rectangle  = length × breadth)  

( Area of triangle =  

2

1

​  

×  Base × Height )  

=t×u+  

2

1

​  

×t×(v−u)

(from the first equation of motion v−u=at)

=ut+  

2

1

​  

×t×at

Thus S=ut+  

2

1

​  

at  

2

......(ii)

This is the second equation of uniform accelerated motion.

(iii) Third equation of motion  : Distance travelled by the object in time interval t is s = area of trapezium OSQP

=  

2

1

​  

(OP+SQ)×OS

∵OP=SR

=  

2

1

​  

(SR+SQ)×OS........(iii)

Acceleration , a  = slope of the velocity -time graph PQ

or a=  

PR

RQ

​  

=  

OS

SQ−SR

​  

 

or OS=  

a

SQ−SR

​  

...........(iv)

Putting this value in equation (iii) we get

s=  

2

1

​  

(SR+SQ)(  

a

SQ−SR

​  

)

or s=  

2a

1

​  

(AQ  

2

−SR  

2

)

or s=  

2a

1

​  

(v  

2

−u  

2

)

or v  

2

−u  

2

=2as

or v  

2

=u  

2

+2as.............(v)  

This the third equation of uniform accelerated motion.

Explanation:

For clarification regarding Graphical Method:-- https://www.toppr.com/en-in/ask/question/derive-the-equations-of-motion-with-the-help-of-the-graphical-method/

Answer 2

Answer

Derivation of First Equation of Motion by Graphical Method

The first equation of motion can be derived using a velocity-time graph for a moving object with an initial velocity of u, final velocity v, and acceleration a.

Derivation of Equation of Motion

In the above graph,

The velocity of the body changes from A to B in time t at a uniform rate.

BC is the final velocity and OC is the total time t.

A perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).

Following details are obtained from the graph above:

The initial velocity of the body, u = OA

The final velocity of the body, v = BC

From the graph, we know that

BC = BD + DC

Therefore, v = BD + DC

v = BD + OA (since DC = OA)

Finally,

v = BD + u (since OA = u) (Equation 1)

Now, since the slope of a velocity-time graph is equal to acceleration a,

So,

a = slope of line AB

a = BD/AD

Since AD = AC = t, the above equation becomes:

BD = at (Equation 2)

Now, combining Equation 1 & 2, the following is obtained:

v = u + at

Derivation of First Equation of Motion by Calculus Method

Since acceleration is the rate of change of velocity, it can be mathematically written as:

a=dvdt

Rearranging the above equation, we get

adt=dv

Integrating both the sides, we get

∫t0adt=∫vudv

at=v−u

Rearranging, we get

v=u+at

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