Question

Velocity of car accelerating from rest in straight line. Equation v(t) = A (1-e ^ -t/t max speed) ^ means to the power of. v(t) is the instantaneous velocity of the car (m/s), t is time in seconds, tmaxspeed is time to reach the maximum speed in seconds and A is a constant. We are given the information that the 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 velocity vs time? 2. 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 this function as t ⮕ ∞?

Answer 1

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

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

$\int ^{x}_{400}dx=\int ^{t}_{10.46} A(1-e^{-t/t_o}) dt$

$x-400=A((t-10.46)-(e^{-t/t_o}-e^{-10.46/t_o}) dt$

$x=-90.322(t-10.46-e^{-t/t_o}+e^{-10.46/t_o})+400$

When t is 0:

$x=-90.322(-10.46-1+e^{-10.46/7})+400=1414.82m$

When t tends to infinity; x tends to infinity.

No asymptote.