Question

the position of a particle in x axis is given by x=t^3-t^2+2t find the velocity when time is 2secs

Answer 1

GOOD MORNING!!

Equation of Trajectory of particle is given as

x = t³ - t² + 2t

Instantaneous velocity can be calculated by its derivative with respect to t

dx/dt = 3t² - 2t + 2

V = 3t² - 2t + 2

where V is instantaneous velocity of the particle.

Velocity At t = 2sec can be calculated as

V = 3 (2)² - 2 (2) + 2

V = 12 - 2

V = 10m/s

So, instantaneous velocity of the particle At t = 2sec is 10m/s

Answer 2

given that the position of the particle in x axis is -

$x \: = {t}^{3} - </u></em></strong><strong><em><u> {t}^{2} + 2t$

$and \: time \: (t) = 2sec$

$now \: \: to \: find \: out \: the \\ \\\: velocity \: of \: the \: particle \: we \: will \: have \\ \\ \: to \: differentiate \: ( x .dt) \: it \:$

(since velocity is differentiation of position)

$now \: \frac{d}{dt} ( {t}^{3} - {t}^{2} + 2t) \\ \\ \\ = 3 {t}^{2} - </u></em></strong><strong><em><u>2</u></em></strong><strong><em><u>t \: + </u></em></strong><strong><em><u>2</u></em></strong><strong><em><u>\\ \\ now \: as \: t = 2 \: (given ) \: put \: value \\ \\ v \: = 3 ({2}^{2} ) - 2 \\ \\ = 12 -2</u></em></strong><strong><em><u>(</u></em></strong><strong><em><u>2</u></em></strong><strong><em><u>)</u></em></strong><strong><em><u>+</u></em></strong><strong><em><u>2</u></em></strong><strong><em><u>\\ \\ \\ = </u></em></strong><strong><em><u>1</u></em></strong><strong><em><u>0$

thus the velocity is 10 m/s

hope helped..