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 acceleration 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

$v(t)=A(1-e^{-t/t_{max}})$

$v(0)=0, \ \ v(2.6)=28\ m/s$

$t_{max}=7\ s$

Find $a(t), a (0)$ and $a(t),t\rightarrow \infin$

Solution:

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

$v(t)=90.25(1-e^{-t/7})$

$a(t)=\frac{dv}{dt}=-90.25e^{-t/7}\times \frac{-1}{7}=12.9 e^{-t/7}$

$a(0)=12.9 \ m/s^2$

$\lim\limits_{t\rightarrow \infin} a(t)= \lim\limits_{t\rightarrow \infin} 12.9e^{-t/7}=0\Rightarrow y=0$  is an horizontal asymptote.

Answer: $a(t)=12.9e^{-t/7}, \ a(0)=12.9 \ m/s^2, \ y=0.$

Information that $x(10.46)=400\ m$ (where $x(t)$ is position of the car as a function of time), can be used to find formula for $x(t)$ :

$\frac{dx}{dt}=v, \ \ x=\int v \ dt= \int (90.25-90.25e^{-t/7})dt= c+90.25t+7\times 90.25e^{-t/7}$

$x(10.46)=400=c+90.25\times 10.46+7\times 90.25e^{-10.46/7} \Rightarrow c=-685.8$

$x(t)=-685.8+ 90.25t+631.75e^{-t/7}$