Question

Derive the second equation of motion using graphical method​

Answer 1

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Answer:

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.

Step-by-step explanation:

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

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Answer 2

Answer:

$\large \underline \boldsymbol{✰ \: \: According \: \: to \: \: the \: \: question : }$

$\qquad \bullet \: \: {\underline{ \tt{We \: have \: to \: \bf \: DERIVATE \: \: \sf Second \: equation \: of \: motion , \: \tt \: by \: \bf \: graphical \: method.}}}$

$\large \underline \boldsymbol{✰ \: \:Answer \: \: of \: \: the \: \: question : }$

$\underline{ \boxed{ : \implies \: { \purple{ \: \frak{ s \: = \: ut \: + \: \frac{1}{2} \: at {}^{2} }}}}}$