Question

Draw velocity –time graph for a body that has initial velocity ‘u’and is moving with uniform acceleration ’a’. Use it to derive v=u+ at, s= ut +1/2at2
plz do step by step
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Answer 1

Answer:

1. v=u+at

2. s=ut+1/2at²

Explanation:

1... final velocity= initial velocity +acceleration ×time

for such a body there will be an acceleration

a= change in velocity /time taken

a=OB-OA/OC

a=v-u/t

a×t=v-u

u+at=v

v=u+at

2... Area of rectangle+ area of triangle

s= l×b+1/2×b×h

OC×OA+1/2×AD×BD

s=t×u+1/2×t×at ( from first equation)

s=ut+1/2at²

Answer 2

v= u+ at

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.



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