Question

derive the 2nd equation of motion by graphically

Answer 1

distance travelled by the body= area under the graph.

s=1/2 multiply by base multiply by height + L multiply by B.

s= 1/2 multiply by t multiply by (v-u) +t multiply by v.

s= (v-u) t /2 + ut.

s= at multiply by t /2 + ut.

s= 1/2 at square + ut

OR

S= UT + 1/2 AT SQUARE

Answer 2

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$\huge\bold\green{ANSWER:}$

✿ Deriving the second equation of motion (Position-time relation)- graphically:-

s= ut + 1/2 at²

As per the graph,

OA= u= initial velocity

OE= v= final velocity

OC= t= time taken

Let the acceleration of the body changes at a uniform rate and it covers a displacement of 's' in time..

Now, from the graph,

→s= Area under graph AB

→s= Area under trapezium OABC

→s=Area of triangle ABD+ Area of the rectangle OADC

Therefore, s= 1/2×b×h + l×b

→s= 1/2× AD×BD + OC×OA

→s=1/2× AD ×(BC-DC)+ OC×OA

Now, substitute the values from the graph..

→s= 1/2 × t × (v-u) + t × u

→s= 1/2 × (v-u)t + ut [Consider it as the equation- 1..]

From the 1st equation of motion, we know that v= u+at or at= v-u [Consider it as the equation -2]

On substituting equation 2 in equation 1, we get,

→s= 1/2 × (v-u)t +ut

→s= 1/2× (at)t + ut

→s= 1/2× at² +ut [Reorder the terms..]

→s= ut + 1/2 at² [Second equation of motion..]

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Hope it helps...:-)

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