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

Answer:

$c=9\ m/s^2$

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, $v=\dfrac{dx}{dt}$

$v=\dfrac{d(3m+4t+ct^2-2t^3)}{dt}$

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

Acceleration, $a=\dfrac{dv}{dt}$

$a=\dfrac{d(4+2ct-6t^2)}{dt}$  

$a=2c-12t$

At t = 3 s

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

Force acting on the particle is given by :

F = ma

$-36\ N=2\ kg\times (2c-36)$

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

Hence, this is the required solution.

Answer 2

Answer:

$c = 9\,\frac{m}{s^{2}}$

Explanation:

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

$a_{x} = -\frac{36\,N}{2\,kg}$

$a_{x} = -18\,\frac{m}{s^{2}}$

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

$v_{x} = 4\,\frac{m}{s} + 2\cdot c \cdot t - \left(6\,\frac{m}{s^{3}} \right)\cdot t^{2}$

$a_{x} = 2 \cdot c - \left(12\,\frac{m}{s^{3}}\right)\cdot t$

The value of c is:

$c = \frac{a_{x}+\left(12\,\frac{m}{s^{3}}\right)\cdot t}{2}$

$c = \frac{-18\,\frac{m}{s^{2}}+\left(12\,\frac{m}{s^{3}} \right)\cdot (3\,s)}{2}$

$c = 9\,\frac{m}{s^{2}}$