Answer to Question #103808 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. Apply the mathematical model above to the car, use given data for the 0-28m/s and 400m times to calculate: value of the coefficient A, maximum velocity and maximum acceleration? please show graphs for this information aswell

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

v(t) is ascending function, so maximum velocity will be on right end, i.e. with $t \to \infty:$

$v_{max}=v(t \to \infty)=90.25m/s$

acceleration: $a(t)=dv/dt=90.25/7*e^{-t/7}$

a(t) is descending function, so the maximum acceleration will be on left end, i.e. with t=0:

$a_{max}=a(t=0)=90.25/7=12.9m/s^2$