Question

A 2 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by x = 3 m+(4 m/s)t+ct2 – (2 m/s3)t3, with x in meters and t in seconds. Factor c is a constant. At t = 3 s, the force on the particle has a magnitude of 36 N and is in the negative direction of the axis. What is c?

Answer 1

`c = 9\,(m)/(s^(2))`

Step-by-step explanation:

The acceleration experimented by the 2 kg particle at t = 3 s. is:

`a_(x) = -(36\,N)/(2\,kg)`

`a_(x) = -18\,(m)/(s^(2))`

The acceleration function is found by differentiating the position function twice:

`v_(x) = 4\,(m)/(s) + 2\cdot c \cdot t - \left(6\,(m)/(s^(3)) \right)\cdot t^(2)`

`a_(x) = 2 \cdot c - \left(12\,(m)/(s^(3))\right)\cdot t`

The value of c is:

`c = (a_(x)+\left(12\,(m)/(s^(3))\right)\cdot t)/(2)`

`c = (-18\,(m)/(s^(2))+\left(12\,(m)/(s^(3)) \right)\cdot (3\,s))/(2)`

`c = 9\,(m)/(s^(2))`

Answer 2

`c=9\ m/s^2`

Step-by-step explanation:

It is given that,

Mass of the particle, m = 2 kg

Force acting on the particle, F = -36 N (negative axis)

The position of the particle as a function of time t is given by :

`x=3m+4t+ct^2-2t^3`

Velocity,
`v=(dx)/(dt)`

`v=(d(3m+4t+ct^2-2t^3))/(dt)`

`v=4+2ct-6t^2`

Acceleration,
`a=(dv)/(dt)`

`a=(d(4+2ct-6t^2))/(dt)`

`a=2c-12t`

At t = 3 s

`a=(2c-36)\ m/s^2`

Force acting on the particle is given by :

F = ma

`-36\ N=2\ kg* (2c-36)`

On solving above equation,
`c=9\ m/s^2`. It has a unit same as acceleration.

Hence, this is the required solution.