Question

Will mark as BRAINLIEST....... The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation 4x³+3x²-5x+2 , where x is in meters and t is in sec. a)Find velocity of particle at i) t=2 sec ii) t=4 sec. b) Find the acceleration of the particle at t=3 sec.

Answer 1

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

$x=4t^3+3t^2-5t+2$

Where,

x is in meters and t is in sec

We know that,

Velocity,

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5$

(a) i. t = 2 s

$v=12(2)^2+6(2)-5=55\ m/s$

At t = 4 s

$v=12(4)^2+6(4)-5=211\ m/s$

(b) Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6$

Pu t = 3 s in the above equation

So,

$a=24(3)+6\\\\a=78\ m/s^2$

Hence, this is the required solution.