Question

The motion of a particle in a straight line is defined by the relation x=t^4-12t^2-40 where x is in meters and t is in sec. Determine the position of x ,velocity v and acceleration a of the particle at t=2sec.​

Answer 1

Answer:

$\red {QUESTION}$

The motion of a particle in a straight line is defined by the relation$x = {t}^{4} - {12t}^{2} - 40$where x is in meters and t is in sec. Determine the position of x ,velocity v and acceleration a of the particle at t=2sec.

$\green {ANSWER}$

$x = {t}^{4} - {12t}^{2} - 40$

At t= 2sec

$x = {(2)}^{4} - {12(2)}^{2} - 40$

$x = - 72m$

Hence position or displacement = -72m

By differentiation method:-

$velocity(v) = d(x)</em><em>/</em><em>dt </em><em>\</em><em>\</em><em> </em><em>= d( {t}^{4} - {12t}^{2} - 40)</em><em>/</em><em>dt$

$= {4t}^{3} - 24t$

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Answer 2

Answer:

At t=2sec position x=(2^4-12*2^2-40)=-80 (-ve sign indicates particle moving in -ve x direction)

velocity v=dx/dt=4*t^3-24*t

at t=2sec v=4*2^3-24*2=-16m/sec

acceleration=dv/dt=12*t^2-24

at t=2sec a=12*2^2-24=24m/sec^2