Answer:
Although this question is not complete, I would give a general solution to this kind of problems.
If y(t) describes the position of a body with time such that
y(t) = at^(n) + bt^(m) + C
Then
V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)
Explanation:
As an example supplies the position of a particle is given by
y(t) = 4t³- 3t² + 9
V(t) = 4x3t²- 3x2t¹
V(t) = d(t)/dt = 12t² - 6t.
Another example,
If y(t) = 15t³ - 2t² + 30t -80
V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.
Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.
One last example,
y(t) = 2t^6 -3t²
V(t) = d(t)/dt = 12t^5 - 6t
Explanation:
a)
θ = 4.91 + 9.7t + 2.06t² when t = 0
θ = 4.91 rad
θ = 4.91 + 9.7t + 2.06t²
ω = dθ/dt = 9.7 + 2.06t, when t =0
ω = dθ/dt = 9.7 + 0
ω = 9.7 rad/s
α = d²θ/dt² = 2.06
α= 2.06 rad/s²
b) please use same method above for t = 2.94 s
Jupiter has a giant red spot
Answer:
where is option
you only say choose that following select apply
Explanation:
please give option
