Question

Question No. 117 :

Deduce the expressions for velocity and acceleration of a particle moving in a straight line.

Answer 1

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Answer 2

Answer:

$\large{\boxed{\sf{Using\:Calculus\:Method-}}}$

1). Derivation of velocity - time relation :

We know that, Acceleration is nothing but the rate of change of velocity.

i.e, $\large{\sf{a = \dfrac{dv}{dt}}}$

$\longrightarrow$ $\large{\sf{dv = adt}}$

Integrating on both sides,

$\longrightarrow$ $\displaystyle\int dv = a \displaystyle\int dt$

As acceleration is constant, so we take it outside the integral. On velocity, we take limit u to v and time from 0 to t.

$\longrightarrow$ $\displaystyle\int\limits^{v}_{u} dv = a \displaystyle\int\limits^{t}_{0} dt$

$\longrightarrow$ $\large{\sf{v\bigg|\limits^{v}_{u} = at}}$

$\longrightarrow$ $\large{\sf{v - u = at}}$

$\longrightarrow$ $\large{\sf{\blue{v = u + at}}}$

2). Derivation of displacement - time relation :

Velocity is rate of change of position with respect to time.

$\longrightarrow$ $\large{\sf{v = \dfrac{dx}{dt}}}$

$\longrightarrow$ $\large{\sf{vdt = dx}}$

Here, "v" isn't independent of "t".

Let's replace v by u + at.

$\longrightarrow$ $\large{\sf{(u + at)dt = dx}}$

Integrating,

$\longrightarrow$ $u \displaystyle\int\limits^{t}_{0} dt + a \displaystyle\int\limits^{t}_{0} tdt = \displaystyle\int\limits^{x}_{x_0} dx$

$\longrightarrow$ $\large{\sf{ut + \dfrac{1}{2} at^{2} = (x - x_{0})}}$

$\longrightarrow$ $\large{\sf{\blue{x = x_{0} + ut + \dfrac{1}{2} at^{2}}}}$

3). Derivation of velocity - displacement relation :

We can write,

$\large{\sf{a = \dfrac{dv}{dt}}}$

Multiplying and dividing by dx,

$\longrightarrow$ $\large{\sf{a = \dfrac{dv}{dx} \times \dfrac{dx}{dt}}}$

$\longrightarrow$ $\large{\sf{a = v \dfrac{dv}{dx}}}$

$\longrightarrow$ $\large{\sf{adx = vdt}}$

Integrating,

$\longrightarrow$ $a \displaystyle\int\limits^{x}_{x_0} dx = \displaystyle\int\limits^{v}_{u} vdt$

$\longrightarrow$ $\large{\sf{a(x - x_{0}) = \dfrac{v^{2} - u^{2}}{2}}}$

$\longrightarrow$ $\large{\sf{\blue{v^{2} - u^{2} = 2a (x - x_{0})}}}$

Here, $\bold{(x - x_{0})}$ is displacement.