Answer to Question #101089 in Calculus for Leo

Question

Velocity of car accelerating from rest in straight line. Equation, v(t) = A (1-e ^ -t/t max speed) the ^ means to power of. v(t) is instantaneous velocity of the car (m/s), t is time in seconds, tmaxspeed is time to reach maximum speed in seconds, A is a constant. We are given the information that car has a t(0-28m/s) of 2.6s, a t(400m) of 10.46s, and a tmaxspeed of 7s (S stands for seconds in this instance). The coefficient A are said to be a unit same as the velocity, that is, m/s. The physical meaning of A is Terminal Velocity(as t tends to infinity, velocity tends to A). The velocity of the car at t=0 and 4 is v(t) = A (1-e ^ -t/t max speed), A(1-e ^ -(2.6/7) ) = -28, A(0.31) = -28, A= -28/0.31 = -90.322m/s, v (0) = A(1-1) = 0 m/s. The asymptote of this function as t ⮕ ∞ is y = -90.322. 1. Sketch a graph of position vs time? 2. 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 this function as t ⮕ ∞?

Answer 1

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

$dv/dt=A(e^{-t/t_o})/t_o$

$a(t)=A(e^{-t/t_o})/t_o$

When t is 0:

$a(0)=-90.322/7=-12.903 m/s^2$

When t tends to infinity; a(t) tends to 0.

Asymptote: $y=0$

1.