Answer to Question #346106 in Calculus for israa

Question

1. Apply your mathematical models to your allocated car. Use the given data for the 0 – 28 m/s and 400m times to calculate the:

1. value of the coefficient A

2. maximum velocity

3. maximum acceleration.

Given: t (0-28 m/s) is 1.9s, t (400m) is 10.50s & tmaxspeed is 7.1s.

Given velocity equation :V (t) = A (1 - e^((-t)/(t maxspeed)))

i need step by step solution to understand

Answer 1

 

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

Using the information that $t_{maxspeed}=7.1\ s$ we have

$v(t) = A (1-e ^ {- t/7.1})$

Using the information that $t (0-28 m/s)$ is $1.9\ s$ we have

$v(1.9) = A (1-e ^ {- 1.9/7.1})=28$

$A=\dfrac{28}{1-e ^ {- 1.9/7.1}}m/s$

$A=119.255\ m/s$

2.

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

$v_{max}=A(1-e^{-1})$

$v_{max}=119.255\ m/s\cdot(1-e^{-1})$

$v_{max}=75.387\ m/s$

3.

$a(t)=\dfrac{dv}{dt}=\dfrac{A}{t_{maxspeed}} (e ^ {- t/t_{maxspeed}})$

$=\dfrac{119.255\ m/s}{7.1\ s} (e ^ {- t/7.1}), 0\ s\le t\le7.1\ s$

The function $a(t)$ decreases for $0\ s\le t\le7.1\ s.$

$a_{max}=a(0)=\dfrac{119.255\ m/s}{7.1\ s}$

$a_{max}=16.797\ m/s^2$