Question

For a graph shape of straight line sloping up starting from origin, between “distance” and “time”, use 2nd equation of motion to deduce the relationship “distance α time” (Hint: uniform motion has no acceleration but has constant velocity)

Answer 1

Given:

For a graph shape of straight line sloping up starting from origin, between “distance” and “time”.

To find:

Deduce the relationship “distance and time”.

Calculation:

Applying 2nd Equation of Motion with uniform acceleration a :

$\therefore \: s = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: s = ut + \dfrac{1}{2} \times \bigg( \dfrac{v - u}{t} \bigg) \times {t}^{2}$

$= > \: s = ut + \dfrac{1}{2} \times \bigg( \dfrac{v - u}{ \cancel{t}} \bigg) \times {t}^{ \cancel{2}}$

$= > \: s = ut + \bigg( \dfrac{v - u}{ 2} \bigg) \times t$

$= > \: s = \bigg(u+ \dfrac{v - u}{ 2} \bigg) \times t$

$= > \: s = \bigg( \dfrac{2u + v - u}{ 2} \bigg) \times t$

$= > \: s = \bigg( \dfrac{u + v }{ 2} \bigg) \times t$

But for uniform velocity , Acceleration is zero.

$\therefore \: s = ut + \dfrac{1}{2} a {t}^{2}$

$=> \: s = ut + \dfrac{1}{2} \times(0)\times {t}^{2}$

$=> \: s = ut$

$=> \: s \:\propto\: t$

Hence , the graph between Displacement and Time will be a straight line passing through origin and having positive slope