+The motion of a particle, along X-axis, is given by theequation x = 9

1.	+The motion of a particle, along X-axis, is given by theequation x = 9 + 5t^2 where 'x' is the distance in metre and't' is time in second,find the instantaneous velocity aftert = 3 s.
Answer» Answer: The instantaneous velocity after 3 SECS is 30 m/s Explanation: $\rule{300}{1.5}$ It's GIVEN that the motion of a particle is given by the equation x = 9 + 5t², here we need to find the instantaneous velocity of the particle at t = 3 seconds. so, from the RELATION between instantaneous velocity and DISPLACEMENT we know that, $\\$ $\longrightarrow\large{\boxed{\sf v = \dfrac{dx}{dt}}}$ Here, v denotes instantaneous velocity. dx denotes small displacement. dt represents short interval of time. $\\$ Substituting the values, $\longrightarrow\sf v=\dfrac{d\bigg(9+5\;t^{2}\bigg)}{dt}\\\\\\\\\longrightarrow\sf v=\dfrac{d\;\Big(9\Big)}{dt}+\dfrac{d\;\Big(5t^{2}\Big)}{dt}\\\\\\\\\longrightarrow\sf v=0+\bigg(2\times 5\bigg)\times t\\\\\\\\\longrightarrow\sf v=0+10\;t\\\\\\\\\longrightarrow\sf v=10\;t$ Substituting t = 3 seconds $\longrightarrow\sf v = 10\times 3\\\\\\\\\longrightarrow\sf v = 30 \\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf v=30\;m/s}}}}$ $\\$ ∴ The instantaneous velocity after 3 secs is 30 m/s. $\rule{300}{1.5}$

+The motion of a particle, along X-axis, is given by theequation x = 9 + 5t^2 where 'x' is the distance in metre and't' is time in second,find the instantaneous velocity aftert = 3 s.

Answer»

Answer:

The instantaneous velocity after 3 SECS is 30 m/s

Explanation:

$\rule{300}{1.5}$

It's GIVEN that the motion of a particle is given by the equation x = 9 + 5t², here we need to find the instantaneous velocity of the particle at t = 3 seconds. so, from the RELATION between instantaneous velocity and DISPLACEMENT we know that,

$\\$

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

Here,

v denotes instantaneous velocity.
dx denotes small displacement.
dt represents short interval of time.

$\\$

Substituting the values,

$\longrightarrow\sf v=\dfrac{d\bigg(9+5\;t^{2}\bigg)}{dt}\\\\\\\\\longrightarrow\sf v=\dfrac{d\;\Big(9\Big)}{dt}+\dfrac{d\;\Big(5t^{2}\Big)}{dt}\\\\\\\\\longrightarrow\sf v=0+\bigg(2\times 5\bigg)\times t\\\\\\\\\longrightarrow\sf v=0+10\;t\\\\\\\\\longrightarrow\sf v=10\;t$

Substituting t = 3 seconds

$\longrightarrow\sf v = 10\times 3\\\\\\\\\longrightarrow\sf v = 30 \\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf v=30\;m/s}}}}$

$\\$

∴ The instantaneous velocity after 3 secs is 30 m/s.

$\rule{300}{1.5}$