Answer:
[tex]c=9\ m/s^2[/tex]
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 :
[tex]x=3m+4t+ct^2-2t^3[/tex]
Velocity, [tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(3m+4t+ct^2-2t^3)}{dt}[/tex]
[tex]v=4+2ct-6t^2[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d(4+2ct-6t^2)}{dt}[/tex]
[tex]a=2c-12t[/tex]
At t = 3 s
[tex]a=(2c-36)\ m/s^2[/tex]
Force acting on the particle is given by :
F = ma
[tex]-36\ N=2\ kg\times (2c-36)[/tex]
On solving above equation, [tex]c=9\ m/s^2[/tex]. It has a unit same as acceleration.
Hence, this is the required solution.
Answer:
[tex]c = 9\,\frac{m}{s^{2}}[/tex]
Explanation:
The acceleration experimented by the 2 kg particle at t = 3 s. is:
[tex]a_{x} = -\frac{36\,N}{2\,kg}[/tex]
[tex]a_{x} = -18\,\frac{m}{s^{2}}[/tex]
The acceleration function is found by differentiating the position function twice:
[tex]v_{x} = 4\,\frac{m}{s} + 2\cdot c \cdot t - \left(6\,\frac{m}{s^{3}} \right)\cdot t^{2}[/tex]
[tex]a_{x} = 2 \cdot c - \left(12\,\frac{m}{s^{3}}\right)\cdot t[/tex]
The value of c is:
[tex]c = \frac{a_{x}+\left(12\,\frac{m}{s^{3}}\right)\cdot t}{2}[/tex]
[tex]c = \frac{-18\,\frac{m}{s^{2}}+\left(12\,\frac{m}{s^{3}} \right)\cdot (3\,s)}{2}[/tex]
[tex]c = 9\,\frac{m}{s^{2}}[/tex]
