Respuesta :
Answer:
a.[tex]\rm -1.49\ m/s^2.[/tex]
b. [tex]\rm 50.49\ m.[/tex]
Explanation:
Given:
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
(a):
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
[tex]\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).[/tex]
At time t = 3 seconds,
[tex]\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.[/tex]
Note: The arguments of the sine is calculated in unit of radian and not in degree.
(b):
The velocity of the particle at some is defined as the rate of change of the position of the particle.
[tex]\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.[/tex]
For the time interval of 2 seconds,
[tex]\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt[/tex]
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
[tex]\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.[/tex]
It is the displacement of the particle in 2 seconds.
