Answer to Question #190543 in Calculus for Daniel

Question

Given the equation,

v(t) = 3 Cos(πt) − 2Sin(πt) (Eq. 1)

● v(t) is the instantaneous velocity of the car (m/s)

● t is the time in seconds

Your tasks is

a) Derive an equation x(t) for the instantaneous position of the particle as

a function of time using indefinite integration.

b) Sketch a graph of position vs. time over the time interval 0 ≤ t ≤ 3

seconds for Eq.1, where C=12.

c) Find a mathematical model (e.g. equation) to correlate position and

time using an Excel sheet and trendline.

d) Using definite integration and driven equation (from c) to find the

position over the time interval 0 ≤ t ≤ 3 seconds and C=12.

e) Using a mid-ordinate rule and driven equation (from c) to find the

position over the time interval 0 ≤ t ≤ 3 seconds at h= 0.5.

f)Find an accurate mathematical model (e.g. equation) to correlate

position and time. To complete this task you should be able to sketch

the graph again, find the accurate equation using an excel sheet and

trendline.

Answer 1

(a) $x(t)=\int v(t)dt$

$=\int (3cos(\pi t)-2sin (\pi t))dt\\[9pt] =-3\pi sin\pi t-2\pi cos (\pi t)$

(b)

(c)The calculus method that can be used here is by putting V=$\dfrac{dx}{dt}$ ,and than by integrating both side of the equation we can get instantaneous equation of a particle covered by the particle during the whole journey and the mathematical model that can be used is that we can put t=0 and can get instantaneous distance of particle at any time.

(D)$x(t)=\int_0^3v(t)dt$

$=\int_0^3 (3cos(\pi t)-2sin (\pi t))dt\\[9pt] =-3\pi sin\pi t-2\pi cos (\pi t)|_0^3\\[9pt]=-3\pi sin3\pi-2\pi cos3\pi-[-3\pi sin0-2\pi cos0]\\[9pt]=-2\pi (-1)+2\pi (1)\\[9pt]=4\pi$

(e)

$x(t)=\dfrac{h}{2}(V(t)-v(0))=\dfrac{0.5}{2}(v(3)-v(0))\\[9pt] =0.25(3cos(3\pi-2sin 3\pi-3cos(0) +2sin(0)) \\[9pt] =0.25\times [3(-1)-3(1)]\\[9pt]=0.25\times -6=-1.5m$

(f) By putting V=$\dfrac{dx}{dt}$ and integrating on both side we can get instantaneous distance of car and by $\dfrac{dv}{dt}$ we can get instantaneous acceleration of car .

Instantaneous distance Equation of car can be given as $x(t)=\dfrac{3sin({\pi}t)}{\pi}+\dfrac{2cos({\pi}t)}{\pi}$

instantaneous acceleration of car can be given as a(t)=$-3{\pi}sin({\pi}t)-2{\pi}cos({\pi}t).$