Question

First law of motion, second law of motion and third law of motion by graphical method.

Answer 1

■HEY DEAR

1 LAW OF MOTION

Slop = teno = perpendicular/ base

a= BD / AB

a= DC - BC/ AB

a=V-U/t

at= V-U

V= U + at.

2 LAW OF MOTION

Area of rectangle OADC = OA× OC

=U × t

=Ut

Area of triangle ABD= 1/2 × Area of rectangle AEBD

=》1/2×AD×BD

=》1/2× t × at

=》1/2at^2

Now,

DISTANCE (S)= sum of the Area of rectangle OADC & Area of triangle ABD

S= Ut + 1/2at^2

3 LAW OF MOTION

Distance travelled = Area of trapezium OABC.

S= (sum of the parallel side) h/2

S= OA + BC × OC / 2

S= (u+ v )×t / 2

Now, to eliminate t

V= U+ at

at= u-v

t= u-v/a

NOW, PUTTING THE VALUE.

S= U+V/2 × U- V/a

S= (U+V) (U-V) / 2a

2as= V^2- U^2

V^2= (-U^2- 2as)

V^2= U^2+2as.

■HOPE ITS HELPFULL

◇◆◇BE BRAINLY◇◆◇

Answer 2

1 LAW OF MOTION

Slop = teno = perpendicular/ base

a= BD / AB

a= DC - BC/ AB

a=V-U/t

at= V-U

V= U + at.

2 LAW OF MOTION

Area of rectangle OADC = OA× OC

=U × t

=Ut

Area of triangle ABD= 1/2 × Area of rectangle AEBD

=》1/2×AD×BD

=》1/2× t × at

=》1/2at^2

Now,

DISTANCE (S)= sum of the Area of rectangle OADC & Area of triangle ABD

S= Ut + 1/2at^2

3 LAW OF MOTION

Distance travelled = Area of trapezium OABC.

S= (sum of the parallel side) h/2

S= OA + BC × OC / 2

S= (u+ v )×t / 2

Now, to eliminate t

V= U+ at

at= u-v

t= u-v/a

NOW, PUTTING THE VALUE.

S= U+V/2 × U- V/a

S= (U+V) (U-V) / 2a

2as= V^2- U^2

V^2= (-U^2- 2as)

V^2= U^2+2as.