Question

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find the displacement at 2 seconds.

Answer 1

a.
`\rm -1.49\ m/s^2.`

b.
`\rm 50.49\ m.`

Step-by-step 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.

`\rm a = (dv)/(dt)\\=(d)/(dt)(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).`

At time t = 3 seconds,

`\rm a=-1.5\sin(0.5* 3)=-1.49\ m/s^2.`

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.

`\rm v = (dr)/(dt).\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.`

For the time interval of 2 seconds,

`\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`

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

`\Delta r=3\ \left ((\sin(0.5\ t))/(0.05) \right )\limits^2_0\\=3\ \left ((\sin(0.5* 2)-sin(0.5* 0))/(0.05) \right )\\=3\ \left ((\sin(1.0))/(0.05) \right )\\=50.49\ m.`

It is the displacement of the particle in 2 seconds.