Numerical solution of trig equation

VSF

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Jul 26, 2012
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I don't have a calculator or math program like Mathematica at hand, so please help me solve this equation for v numerically:

1 = 18tan(v) - 7938/cos2(v)

or

1 = 18sin(v)sec(v)-7938sec2(v)


Thanks a lot in advance!



Background

This problem is based on solving for the velocity of a car that drove off the road in my hometown in Lillestrøm, Norway, killing the two people in the car. The car came out of a tunnel where it passed a speed camera which didn't register the car's speed. The camera has a lower cutoff at 40 km/h and an upper cutoff at 256 km/h so the car must have been driving faster than 256km/h since it would not have had time to accelerate enough if it was going below 40km/h. It is the first time this has happened. The motorway inside the tunnel very cleverly ends in a roundabout just outside the tunnel opening. The car hit the curb of the central circle of the roundabout and was airborne across the fence, down a hill, cutting the tops off some bushes, passed above another fence, by a walking path and was found 15 meters from the edge of a river running at normal angle to the tunnel motorway and flight path. In total the car flew 90 meters and I assume that its center of gravity must have been 5 meters up in the air at the highest point of the flight path in order to pass all those items.

I figured that estimating the initial velocity of the car would be an interesting problem to solve for high school students or higher that are learning about trig functions, and I wanted to check if my solution is valid before presenting the problem to anybody else. Here goes:

Accident info (in Norwegian):
http://www.dagbladet.no/2012/07/26/nyheter/innenriks/bilulykke/relingen/lillestrom/22688668/

Based on the info from the newspaper I assume the following parameters/variables. I choose to solve for the initial airborne velocity of the car at a slight upwards angle after it hit the curb, rather than starting with the horizontal speed along the motorway. That velocity is given by v0motorway = v0airborne tan (5/45) and can easily be calculated after the initial airborne velocity has been determined. Here I disregard the friction when hitting the curb and air resistance, and I use a simple tangent instead of considering the slightly higher angle due to the ballistic flight path.

Analytical solution

I take these as negligible factors for this rough estimate.

g =9,8m/s2
v0 =vm/sv >256km/h71,11111m/s
t =ts
x =90m
y =5m
I have the following basic relationships:

y = v0y*t - 1/2*g*t2
x = v0x*t
v0y = sin(v)
v0x = cos(v)

I substitute parameters and variables and simplify:

5 = v0yt - 4,9t2
and
t = 90/v0x
<=>
5 = 90v0y/v0x - 4,9(90/v0x)2
<=>
5 = 90sin(v)/cos(v) - 4,9*902/cos2(v)
<=>
1 = 18tan(v)-7938/cos2(v)
<=>
1 = 18sin(v)sec(v)-7938sec2(v)

Again, thanks in advance for helping out!
 
Hello, VSF!


Your equation has no solution.


You have: .\(\displaystyle 18\tan v - \dfrac{7938}{\cos^2v} \:=\:1 \quad\Rightarrow\quad 18\tan v - 7938\sec^2v \:=\:1\)

. . \(\displaystyle 18\tan v - 7938(\tan^2v + 1) \:=\:1 \quad\Rightarrow\quad 18\tan v - 7938\tan^2v - 7938 \:=\:1\)


We have a quadratic equation: .\(\displaystyle 7938\tan^2v - 18\tan v + 7939 \:=\:0\)

. . which has no real roots.
 
I don't have a calculator or math program like Mathematica at hand, so please help me solve this equation for v numerically:

1 = 18tan(v) - 7938/cos2(v)

or

1 = 18sin(v)sec(v)-7938sec2(v)


Thanks a lot in advance!



Background

This problem is based on solving for the velocity of a car that drove off the road in my hometown in Lillestrøm, Norway, killing the two people in the car. The car came out of a tunnel where it passed a speed camera which didn't register the car's speed. The camera has a lower cutoff at 40 km/h and an upper cutoff at 256 km/h so the car must have been driving faster than 256km/h since it would not have had time to accelerate enough if it was going below 40km/h. It is the first time this has happened. The motorway inside the tunnel very cleverly ends in a roundabout just outside the tunnel opening. The car hit the curb of the central circle of the roundabout and was airborne across the fence, down a hill, cutting the tops off some bushes, passed above another fence, by a walking path and was found 15 meters from the edge of a river running at normal angle to the tunnel motorway and flight path. In total the car flew 90 meters and I assume that its center of gravity must have been 5 meters up in the air at the highest point of the flight path in order to pass all those items.

I figured that estimating the initial velocity of the car would be an interesting problem to solve for high school students or higher that are learning about trig functions, and I wanted to check if my solution is valid before presenting the problem to anybody else. Here goes:

Accident info (in Norwegian):
http://www.dagbladet.no/2012/07/26/nyheter/innenriks/bilulykke/relingen/lillestrom/22688668/

Based on the info from the newspaper I assume the following parameters/variables. I choose to solve for the initial airborne velocity of the car at a slight upwards angle after it hit the curb, rather than starting with the horizontal speed along the motorway. That velocity is given by v0motorway = v0airborne tan (5/45) and can easily be calculated after the initial airborne velocity has been determined. Here I disregard the friction when hitting the curb and air resistance, and I use a simple tangent instead of considering the slightly higher angle due to the ballistic flight path.

Analytical solution

I take these as negligible factors for this rough estimate.

g =9,8m/s2
v0 =vm/sv >256km/h71,11111m/s
t =ts
x =90m
y =5m
I have the following basic relationships:

y = v0y*t - 1/2*g*t2
x = v0x*t
v0y = sin(v)
v0x = cos(v)

I substitute parameters and variables and simplify:

5 = v0yt - 4,9t2
and
t = 90/v0x
<=>
5 = 90v0y/v0x - 4,9(90/v0x)2
<=>
5 = 90sin(v)/cos(v) - 4,9*902/cos2(v)
<=>
1 = 18tan(v)-7938/cos2(v)
<=>
1 = 18sin(v)sec(v)-7938sec2(v)

Again, thanks in advance for helping out!

1 = 18tan(v) - 7938/cos2(v)

1 = 18* tan(v) - 7938*sec2(v)


1 = 18* tan(v) - 7938*[1 + tan2(v)]

Do you see a quadratic equation? - solve it....
 
A simpler way to go about it

Hello, VSF!
We have a quadratic equation: .\(\displaystyle 7938\tan^2v - 18\tan v + 7939 \:=\:0\)

. . which has no real roots.

Thank you both for answering! I have thought about it meanwhile and think it would be easier to solve a quadratic for t and then calculate v, rather than substituting all the trig functions. It might be possible to get an answer that way. I had written up an almost complete analysis but then my browser crashed and now I don't have time to write it up again. In any case what I did was change the value for y to 4,9m which allows for a nice quadratic formula result on monic form, where the discriminator has to be positive and larger than the linear coefficient in order to get a positive value for the initial vertical speed. Expression from Wikipedia:

5609a15c588ef80c2987ecd242e32dd1.png
 
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