Answer to Question #101194 in Calculus for Leo

Question

to calculate velocity of a car accelerating from rest in straight line. Equation is v(t) = A (1-e ^ -t/t max speed) the ^ means to the power of. v(t) is the instantaneous velocity of the car (m/s), t is the time in seconds, tmaxspeed is the time to reach the maximum speed in seconds and A is a constant. 1. Create a graph of acceleration vs time? 2. Apply the mathematical model which is a car with a t(0-28m/s) of 2.6s, a t(400m) of 10.46s in order to calculate the value of coefficient, maximum velocity and maximum acceleration?

Answer 1

$$\text v(t)=A(1-e^{-{t \over t_{max\_ speed}}})$$

$$\text a(t)=\text v'(t)={A \over t_{max\_ speed}}e^{-{t \over t_{max\_speed}}}$$

$$\text v(t)=A(1-e^{-{t \over t_{max\_ speed}}})$$

$$\text s(t)=\int \text v(t) dt=\int A(1-e^{-{t \over t_{max\_ speed}}})dt=$$

$$=A(t+t_{max\_speed}\cdot e^{-{t \over t_{max\_ speed}}})+C$$

$$\text s(0)=0=>C=-A\cdot t_{max\_speed}$$

$$\text s(t)=A(t-t_{max\_speed}+t_{max\_speed}\cdot e^{-{t \over t_{max\_ speed}}})$$

Given that

$\text v(2.6)=A(1-e^{-{2.6 \over t_{max\_ speed}}})=28$

$\text s(10.46)=$$A(10.46-t_{max\_speed}+t_{max\_speed}\cdot e^{-{10.46 \over t_{max\_ speed}}})=400$

$${10.46-t_{max\_speed}+t_{max\_speed}\cdot e^{-{10.46 \over t_{max\_ speed}}} \over 1-e^{-{2.6 \over t_{max\_ speed}}}}={400 \over 28}$$

$t_{max\_speed}\approx5.834\ s$ or $t_{max\_speed}\approx28.424 \ s$

$A\approx77.864\ m/s$ or $A\approx320.318\ m/s$

$$\text v_{max}=A(1-e^{-{ t_{max\_ speed} \over t_{max\_ speed}}})=A(1-e^{-1})$$

$$\text a_{max}={A \over t_{max\_ speed}}e^{-{t_{max\_speed} \over t_{max\_speed}}}={Ae^{-1} \over t_{max\_ speed}}$$

$t_{max\_speed}\approx5.834\ s,A\approx77.864\ m/s$

$$\text v_{max}=77.864(1-e^{-1})\ m/s\approx49.219\ m/s$$

$$\text a_{max}={77.864\ m/s\cdot e^{-1} \over 5.834\ s}\approx4.910\ m/s^2$$

Or

$t_{max\_speed}\approx28.424\ s,A\approx320.318\ m/s$

$$\text v_{max}=320.318(1-e^{-1})\ m/s\approx202.480\ m/s$$

$$\text a_{max}={320.318\ m/s\cdot e^{-1} \over 28.424\ s}\approx4.146\ m/s^2$$

$202.480\ m/s\approx729\ km/h$

I think, that the car cannot have such speed.