A) f = 1.8 rev/s = 2 Hz
T = 1 / f = 0.55s

B) not really sure..srry

C) T = 2 pi √ ( L / g ) 
0.57 = 2 x 3.14 x √ ( 0.2 / g )

g = 25.5 m/s²

Hope this helps a little at least.. :)

5 0

2 years ago

g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0

Margarita [4]

Answer:

Explanation:

If a particle move with time and expressed according to the formula:

f(t) = 0.01t⁴ − 0.03t³

a) Velocity is the change in motion of the particle with respect to time and it is expressed as;

$v(t) =\frac{d(f(t))}{dt}$

$v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2$

Hence the velocity of the particle at time t is $v(t) = 0.04t^3 - 0.09t^2$

b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:

$v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05$

Hence the velocity after 1second is -0.05

c) The particle is at rest when when the time is zero.

Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second

6 0

1 year ago