Question

two straight line drawn on same displacement time graph and make 30 degree and 60 degree with time axis . which represent greater velocity also find ratio of velocity?

Answer 1

Task. Two straight line drawn on same displacement-time graph and make 30 degree and 60 degree with time axis. Which represent greater velocity also find ratio of velocity?

Solution. We have two graphs of displacements functions $d_{1}(t)$ and $d_{2}(t)$. By assumption both of them are straight lines, so

$d_{1}(t)=k_{1}t+b_{1},\qquad d_{2}(t)=k_{2}t+b_{2}$

for some numbers $k_{1},b_{1},k_{2},b_{2}$. Moreover, the first line constitute 30 degree with time axis, and the second one constitute 60 degree.

It is well-known that in teh equaltion of straight line $d(t)=kt+b$ the coefficient $k$ is equal to tangent of the angle $\alpha$ between the line and time axis:

$k=\tan\alpha.$

Hence for the first line

$k_{1}=\tan 30{}^{\circ}=\frac{1}{\sqrt{3}},$

and for the second line

$k_{2}=\tan 60{}^{\circ}=\sqrt{3}.$

Now notice that the velocity of an object is the time derivative of its displacement:

$v(t)=d^{\prime}(t).$

For the linear function $d(t)=kt+b$ its derivative is equal to the coefficient $k$:

$d^{\prime}(t)=\left(kt+b\right)^{\prime}=k.$

Thus for the first object its velocity is

$v_{1}(t)=k_{1}=\frac{1}{\sqrt{3}}\ m/s,$

and for the second one:

$v_{2}(t)=k_{2}=\sqrt{3}\ m/s.$

Therefore the second velocity is greater the the first one:

$v_{1}<v_{2}.$

Their ratio is

$\frac{v_{2}}{v_{1}}=\frac{\sqrt{3}}{1/\sqrt{3}}=\sqrt{3}\cdot\sqrt{3}=3.$

Answer. $v_{2}=3v_{1}.$