Answer to Question #103807 in Calculus for leo

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 derive an equation a(t) for the instantaneous position of the car as a function of time? Identify the acceleration of the car at t = 0s and asymptote of this function as t → ∞?

Answer 1

$v(t)=A(1−e^{−t/t_{maxspeed}})$

using the information that t (0-28 m/s) is 2.6s & tmaxspeed is 7s

$v(2.6)=28=A(1-e^{-2.6/7}) => A=90.25m/s$

finally $v(t)=90.25(1-e^{-t/7})$

Acceleration: $acc(t)=dv/dt=90.25/7*e^{-t/7}=12.9e^{-t/7}$

$acc(t=0)=12.9m/s^2$

Asymptote of this function as t → ∞:

$lim_{t \to \infty}acc(t)=lim_{t \to \infty}12.9e^{-t/7}=0$ ,

then we have horizontal asymptote as t → ∞ : $y(t)=0.$

Derive an equation a(t) for the instantaneous position of the car as a function of time:

$a(t)=a_0+A*t+A*t_{maxspeed}*e^{-t/t_{maxspeed}}$

using that t(400m) is 10.46s & tmaxspeed is 7s & A=90.25:

$a(10.46)=400=a_0+90.25*10.46+90.25*7*e^{-10.46/7}$

$=> a_0=-685.8m$

finally: $a(t)=-685.8+90.25*t+90.25*7*e^{-t/7}$