The displacement of a particle moving along x-axis is given by X = 18t

1.	The displacement of a particle moving along x-axis is given by X = 18t + 5t.calculate:(i) the instantaneous velocity at t = 2s.(ii) average velocity between t = 2 s and t = 3s.(iii) instantaneous acceleration
Answer» GIVEN The displacement of a particle moving along X-axis is given by x = 18t + 5t² To Find i. Instantaneous velocity at t = 2 s ii. Average velocity b/w t = 2 s and t = 3 s iii. Instantaneous ACCELERATION Solution i. $\rm v=\dfrac{dx}{dt}\\\\\implies \rm v=\dfrac{d}{dt}(18t+5t^2)\\\\\implies \rm v=18+10t$ Instantaneous Velocity at t = 2 s is given by , $\implies \rm v=18+10(2)\\\\\implies \bf \pink{v=38\ m/s}\ \; \bigstar$ ii. Average velocity is given by displacement per change in time $\implies \rm A_{v}=\dfrac{x_2-x_1}{t_2-t_1}$ $\implies \rm A_v=\dfrac{\{18(3)+5(3)^2\}-\{18(2)+5(2)^2\}}{3-2}\\\\$ $\implies \rm A_v=\dfrac{\{54+45\}-\{36+20\}}{1}\\\\\implies \rm A_v=99-56\\\\\implies \bf \green{A_v=43\ m/s}\ \; \bigstar$ iii. Instantaneous acceleration is given by , $\implies \rm a=\dfrac{dv}{dt}\\\\\implies \rm a=\dfrac{d}{dt}(18+10t)\\\\\implies \bf \blue{a=10\ m/s^2}\ \; \bigstar$

The displacement of a particle moving along x-axis is given by X = 18t + 5t.calculate:(i) the instantaneous velocity at t = 2s.(ii) average velocity between t = 2 s and t = 3s.(iii) instantaneous acceleration

Answer»

The displacement of a particle moving along X-axis is given by

To Find

i. Instantaneous velocity at t = 2 s

ii. Average velocity b/w t = 2 s and t = 3 s

iii. Instantaneous ACCELERATION

Solution

$\rm v=\dfrac{dx}{dt}\\\\\implies \rm v=\dfrac{d}{dt}(18t+5t^2)\\\\\implies \rm v=18+10t$

Instantaneous Velocity at t = 2 s is given by ,

$\implies \rm v=18+10(2)\\\\\implies \bf \pink{v=38\ m/s}\ \; \bigstar$

ii.

Average velocity is given by displacement per change in time

$\implies \rm A_{v}=\dfrac{x_2-x_1}{t_2-t_1}$

$\implies \rm A_v=\dfrac{\{18(3)+5(3)^2\}-\{18(2)+5(2)^2\}}{3-2}\\\\$

$\implies \rm A_v=\dfrac{\{54+45\}-\{36+20\}}{1}\\\\\implies \rm A_v=99-56\\\\\implies \bf \green{A_v=43\ m/s}\ \; \bigstar$

iii.

Instantaneous acceleration is given by ,

$\implies \rm a=\dfrac{dv}{dt}\\\\\implies \rm a=\dfrac{d}{dt}(18+10t)\\\\\implies \bf \blue{a=10\ m/s^2}\ \; \bigstar$