Question

Derive second equation of motion.

class -9th

Answer 1

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Explanation:

Consider an object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time t = 0 and t = t are u and v respectively. During time t, let s be the total distance travelled by the object. In uniformly acceleration motion the velocity – time graph of an object is a straight line, inclined to the time axis.

OD = u, OC = v and OE = DA = t.

Let, the Initial velocity of the object = u

Let, the object is moving with uniform acceleration, a

Let, the object reaches at point B after time t, and its final velocity becomes, v.

Draw a line parallel to x-axis DA from point, D from where object starts moving.

Draw another line BA from point B parallel to Y-axis which meets at E at y-axis.

Second Equation of Motion: Distance covered by the object in the given time “t” is given by the area of the trapezium ABDOE.

Let in the given time (t), the distance covered by the moving object = s

The area of trapezium, ABDOE.

Distance (s) = Area of ΔABD + Area of ADOE.

s = ½ x AB x AD + (OD x OE)

s = ½ x DC x AD + (u x t) [∵ AB = DC]

s = ½ x at x t + ut [∵ DC = at]

s = ½ x at x t + ut

s = ut + ½ at².

It is the expression gives the distance covered by the object moving with uniform acceleration.

Answer 2

Solution :

Derivation of Second Equation of Motion by Algebraic Method

We know,

Let,

Initial velocity of the object = u

Final velocity of the object = v

Acceleration = a

Time = t

Distance covered in given time = s

We know that,

$\sf{}Average\ Velocity=\dfrac{v-u}{t}$

Also,

Displacement = Average velocity × Time

or,

$\sf{}s=ut+\dfrac{1}{2}at^2.....(1)$

But, v = u + at_____(2) (from 1st eq. of motion)

After substituting the value of ‘v’ from (2) to (1), we get

$\sf{}s=\dfrac{u+u+at}{2}\times t$

$\sf{}s=\bigg(\dfrac{2u+at}{2}\bigg)\times t$

$\sf{}s=\dfrac{2ut+at^2}{2}$

$\sf{}s=\dfrac{2ut}{2}+\dfrac{at^2}{2}$

$\boxed{\sf{}s=ut+\dfrac{1}{2}at^2}$

The above equation is known as Second equation of motion.

Derivation of Second Equation of Motion by Graphical Method.

Distance covered by the object in the given time ‘t’ is given by the area of the trapezium ABDOE

Let in the given time, t the distance covered by the moving object = s

Area of trapezium, ABDOE

$\sf{}Distance (s) = Area\ of\ \Delta ABD+Area\ of\ ADOE$

$\sf{}s= \dfrac{1}{2}\times AB\times AD+ (OD\timesOE)$

$\sf{}s= \dfrac{1}{2}\times DC\times AD+ (u\times t)\ \ \ \ AB=DC$

$\sf{}s= \dfrac{1}{2}\times at\times t+ ut\ \ \ \ DC=at$

$\sf{}s= \dfrac{1}{2}at^2+ ut$

$\boxed{\sf{}s=ut+\dfrac{1}{2}at^2}$

The above expression gives the distance covered by the object moving with uniform acceleration. This expression is known as second equation of motion.