Answer to Question #101090 in Calculus for Leo

Question

Velocity car accelerating from rest in straight line. Equation is v(t) = A (1-e ^ -t/t max speed) the ^ means to power of. v(t) is instantaneous velocity of car (m/s), t is time in seconds, tmaxspeed is time to reach the max speed in seconds and A is a constant. 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. coefficient A are said to be a unit same as velocity, that is, m/s. physical meaning of A is Terminal Velocity(as t tends to infinity, velocity tends to A). 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 acceleration vs time? 2. Apply mathematical model. Use the given data for the 0-28 m/s and 400m times to calculate the: value of the coefficient A, maximum velocity and maximum acceleration?

Answer 1

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

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

$A(0.31)=-28$

$A=-28/0.31=-90.322m/s$

As t tends to 0, (1-e^-t/7) tends to 0.So,maximum velocity is given by:

$v_{max}=0m/s$ ( Can be obtained from graph attached below,t cannot be negative so max value when t is 0)

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

$a(t)= -90.322(e ^ {-t/t_o})/t_o$

From the graph of acceleration vs time,we can get:

$a_{max}=0m/s^2$as graph converges to 0 as t tends to infinity