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v = u + at

Q1) Prove this with the help of graph.

Q2) Prove this without the help of graph

9th class CBSE

Answer 1

Hi there !!

Thanks for the question :)

Graphical method to prove the second equation of motion is in the attachment :)

The method without graph ,
Down here ⬇️⬇️

We know that ,

$a = \frac{v - u}{t}$

where,
a = acceleration
v = final velocity
u = initial velocity
t = time taken

Let us assume acceleration to be a constant

Therefore,


$a = \frac{v - u}{t} \\ = (a) \times (t) = v - u$
at = v - u

Transposing u to LHS ,
we have,

at + u = v
or
v = u + at

There we have it ! the 1st equation of motion

Answer 2

Hey there !!
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Q1) Refer to the attachment for the graph.

From the graph, initial velocity is u (at point A) and then it increases to v (at point B) in time t.
The velocity changes at a uniform rate a.

The initial velocity is represented by OA, the final velocity is represented by BC and the time is represented by OC.
BC – CD = BD represents the change in velocity in time t.

Now, BC = BD + DC = BD + OA
Substituting BC = v and OA = u
v = BD + u
or
BD = v – u ...(1)

As we know, a = Change in velocity / time taken
= BD/AD = BD/OC

Substituting OC = t
a = BD/ t
or
BD = at ...(2)

Using eqⁿ (1) and (2),
v = u + at
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Q2) Let 'a' be acceleration.

a = Change in velocity / time taken

Let 'u' be the initial velocity and 'v' be the final velocity of the object. And time taken is donated by 't'.

Now, a = ( v – u ) / t

=> at = v – u

=> v = u + at
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Hope my ans.'s satisfactory. (^-^)