Question

you have been given the mathematical model to calculate velocity of a car accelerating from rest in straight line: v(t) = A (1-e ^ - t/tmaxspeed) (anything after the symbol ^ is to the power of) v(t is instantaneous velocity of car (m/s), t is time in seconds, tmaxspeed is time to reach maximum speed in seconds and A is constant. you are given the information that t (0-28 m/s) is 2.6s, t (400m) is 10.46s & tmaxspeed is 7s. please explain and derive an equation x(t) for the instantaneous position of the car as a function of time.....identify the value of x when t=0s ? and asymptote of this function as t → ∞?

Answer 1

Given : $v(t) = A (1-e ^{-t/t_m})$ , where $t_m$ is time to reach maximum speed in seconds and A is constant.

Also given; $v(2.6)=28m/s;t_m=7s$

$\implies v(t) = A (1-e ^{-t/7})$

$\implies v(2.6) = A (1-e ^{-2.6/7})=28$

$\implies A=90.25m/s$

$\implies v(t) =90.25(1-e ^{-t/7})=dx/dt$

Integrating on both sides, we get;

$\implies \int dx=90.25 \int (1-e^{-t/7})dt$

$\implies x(t)=90.25(t+e^{-t/7}/7)+c$

Also given; $x(10.46)=400=90.25(10.46+e^{-10.46/7}/7)+c=964.268+c$

$\implies c=-564.268 m$

$\implies x(t)=90.25(t+e^{-t/7}/7)-564.268$ ---(1)

(1) gives instantaneous position of the car as a function of time.

$x(t=0)=90.25(0+e^0/7)-564.268=90.25/7-564.268=-551.375$

$\implies x(0)=-551.375m$ ----(Answer)

as $t \to \infty, e^{-t} \to 0 \implies x \to \infty$

$k=\lim_{t \to \infty} \frac{x(t)}{t}=90.25,$ $b=\lim_{t \to \infty}(x(t)-kt)=\lim_{t \to \infty}(x(t)-90.25t)=-564.268$

The asymptote of this function is $y(t)=90.25t-564.268$.