
Assuming vertical acceleration of
, the speed after x seconds of falling is 
Position: x = 18t y = 4t - 4.9t²
First derivative: x' = 18 y' = 4 - 9.8t
Second derivative: x'' = 0 y'' = - 9.8
Position vector: P = (18t) i + (4t - 4.9t²) j
Velocity vector: V = (18) i + (4 - 9.8t) j
Acceleration vector A = (- 9.8) j
Answer:
a.
b. 
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
<h2>
(a):</h2>
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

At time t = 3 seconds,

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>
(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.

For the time interval of 2 seconds,

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

It is the displacement of the particle in 2 seconds.
Answer:
I do not know which substance you are referring to, but the freezing point of water is 32°F, or 0°C.
The statement that accurately describes a proper use of eyeglasses is statement A. Using converging lenses to help a nearsighted person by moving the image from in front of the retina to the retina.