Can you help with this word problem.

Elwuero

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May 21, 2019
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A car decelerates such that each second it travels 8% less than the distance it traveled in the previous second. How far does the car travel if it is traveling at 20 meters per second during the first second of deceleration?
 
I think what is meant is that the car's initial velocity is 20 m/s. Let's let the car's position at time \(t=0\) be \(x(0)=0\). The velocity of the car will undergo some kind of exponential decay:

[MATH]v(t)=v_0e^{-kt}[/MATH] where \(0<k\)

Hence:

[MATH]x(t)=\frac{v_0}{k}\left(1-e^{-kt}\right)[/MATH]
Now, we are told:

[MATH]x(t+1)-x(t)=0.92(x(t)-x(t-1))[/MATH]
[MATH]x(t+1)-1.92x(t)+0.92x(t-1)=0[/MATH]
[MATH]\frac{v_0}{k}\left(1-e^{-k(t+1)}\right)-1.92\frac{v_0}{k}\left(1-e^{-kt}\right)+0.92\frac{v_0}{k}\left(1-e^{-k(t-1)}\right)=0[/MATH]
[MATH]\left(1-e^{-k(t+1)}\right)-1.92\left(1-e^{-kt}\right)+0.92\left(1-e^{-k(t-1)}\right)=0[/MATH]
[MATH]e^{-k(t+1)}-1.92e^{-kt}+0.92e^{-k(t-1)}=0[/MATH]
[MATH]0.92e^{2k}-1.92e^{k}+1=0[/MATH]
[MATH](e^{k}-1)(23e^{k}-25)=0[/MATH]
This implies:

[MATH]e^{-k}=\frac{23}{25}[/MATH]
Hence:

[MATH]x(t)=\frac{v_0}{\ln\left(\dfrac{25}{23}\right)}\left(1-\left(\frac{23}{25}\right)^t\right)[/MATH]
Here's a graph of the car's position:

fmh_0052.png

And so the total distance \(T\) the car will travel is given by:

[MATH]T=\lim_{t\to\infty}x(t)=\,?[/MATH]
 
I think what is meant is that the car's initial velocity is 20 m/s. Let's let the car's position at time \(t=0\) be \(x(0)=0\). The velocity of the car will undergo some kind of exponential decay:

[MATH]v(t)=v_0e^{-kt}[/MATH] where \(0<k\)

Hence:

[MATH]x(t)=\frac{v_0}{k}\left(1-e^{-kt}\right)[/MATH]
Now, we are told:

[MATH]x(t+1)-x(t)=0.92(x(t)-x(t-1))[/MATH]
[MATH]x(t+1)-1.92x(t)+0.92x(t-1)=0[/MATH]
[MATH]\frac{v_0}{k}\left(1-e^{-k(t+1)}\right)-1.92\frac{v_0}{k}\left(1-e^{-kt}\right)+0.92\frac{v_0}{k}\left(1-e^{-k(t-1)}\right)=0[/MATH]
[MATH]\left(1-e^{-k(t+1)}\right)-1.92\left(1-e^{-kt}\right)+0.92\left(1-e^{-k(t-1)}\right)=0[/MATH]
[MATH]e^{-k(t+1)}-1.92e^{-kt}+0.92e^{-k(t-1)}=0[/MATH]
[MATH]0.92e^{2k}-1.92e^{k}+1=0[/MATH]
[MATH](e^{k}-1)(23e^{k}-25)=0[/MATH]
This implies:

[MATH]e^{-k}=\frac{23}{25}[/MATH]
Hence:

[MATH]x(t)=\frac{v_0}{\ln\left(\dfrac{25}{23}\right)}\left(1-\left(\frac{23}{25}\right)^t\right)[/MATH]
Here's a graph of the car's position:

View attachment 12219

And so the total distance \(T\) the car will travel is given by:

[MATH]T=\lim_{t\to\infty}x(t)=\,?[/MATH]
Wow, thanks for your help! That was a lot of work. Thanks again.
 
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