Question

explain uniform and non uniform accelaration by velocity time graph​

Answer 1

Acceleration of a body = Rate of change of velocity i.e. slope of the velocity-time graph. The velocity-time graph for a uniformly accelerated body is given by the straight line AB. So, acceleration of the body is equal to the slope of the line AB. Acceleration = Slope of line AB =ADBD=ADBC−DC.

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Answer 2

Answer:

The equation of motions from the velocity-time graph:

The velocity-time graph for a body under uniform acceleration is shown in the figure.

Let initial velocity of the body = u

The final velocity of the body = v

Time is taken by the body = t

Acceleration of the body = a

Derivation of the first equation of motion:

According to the definition,

Acceleration of a body = Rate of change of velocity i.e. slope of the velocity-time graph.

The velocity-time graph for a uniformly accelerated body is given by the straight line AB. So, acceleration of the body is equal to the slope of the line AB.

Acceleration = Slope of line AB =

AD

BD

=

AD

BC−DC

From the velocity time graph,

BC=v,DC=OA=u,AD=OC=t

Then, a⇒

t

v−u

⇒at=v−u⇒v=u+at

⇒ First equation of motion

Derivation of the second equation of motion

Under the uniform acceleration, from the figure, one can write

Distance travelled (s) = Area of trapezium OABC

S = Area of the triangle ABD + Area of rectangle OADC

=

2

1

×AD×BD+OC×OA

Here, AD=OC=t

and BD=BC−DC=v−u

BD=u+at−u=at and OA=u

∴s=

2

1

×t×at+t×u⇒s=ut+

2

1

at

2

This is second equation of motion.

Derivation of third equation of motion

Distance travelled, s = Area of the trapezium OABC

s=

2

1

(Sum of the parallel sides) × Perpendicular distance between the two parallel sides

s=

2

1

×(OA+BC)×OC

s=

2

1

(u+v)×t But a=

t

v−u

⇒t=

a

v−u

⇒s=

2

1

(v+u)×

a

v−u

⇒s=

2a

v

2

−u

2

⇒v

2

−u

2

=2as⇒ Third equation of motion.

Explanation:

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