Hello everyone

 

I made myself a question about a car flying of a cliff.. The problem is: the answer seems pretty unrealistic to me:

 

A car drives, when he's at 20 m of the edge of a cliff with a height of 147.63 m, at a velocity of 65 km/h and at a constant acceleration of 10 ms^-2.

Determine how far the car will land of the edge of the cliff.

 

To do this, I'll use the formula [math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2v^2\cos^2{\theta}}[/math].

 

I will thus need the car's velocity when its on the edge of the cliff:

 

[math]v = v_0 + a\cdot t[/math]

 

[math] x = \frac{at^2}{2}[/math]

 

[math]t^2 = \frac{20\cdot 2}{10}[/math]

 

[math]t = 4s[/math]

 

[math]v=18 ms^{-1} + 4s\cdot 10 ms^{-2}= 38 ms^{-1}[/math]

 

[math]0 = 147.63 + x\cdot\tan{0}-\frac{g\cdot x^2}{2\cdot 38^2\cdot \cos^2{0}}[/math]

 

[math]\frac{-g}{2\cdot 38^2}x^2+147.63=0[/math]

 

[math]x^2=43,461.31 m[/math]

 

[math]x=208.47 m[/math]

 

Is this realistic? (Of course, these formulas don't include air resistance..)

 

Thanks.

 

-Function

 

Is this realistic?

 

 

No, as soon as the car wheels leave the ground the car has no traction cannot accelerate.

Yes, but acceleration isn't used in the 'trajectory' formula..? The only speed-related variable it uses is velocity, and that velocity should be 38 ms^-1, which is the car's velocity at the very edge of the cliff..

Edited November 23, 201312 yr by Function

[math]x = \frac{at^2}{2}[/math]

 

[math]t^2 = \frac{20\cdot 2}{10}[/math]

 

[math]t = 4s[/math]

Unless I'm misreading something, this is not correct. Remember the full equation for x is [math]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/math]. Taking the car's initial position to be 0, and the edge of the cliff to be at 20 m, we thus have [math]20 = \frac{65000}{3600}t + \frac{10}{2}t^{2}[/math]. Ignoring the negative root, this gives us [math]t = \frac{8}{9} \approx 0.889\textnormal{ s}[/math], which makes sense, given that 65 km/hr is already a little over 18 m/s.

Edited November 24, 201312 yr by John

Unless I'm misreading something, this is not correct. Remember the full equation for x is [math]x = x_{0} + v_{0}t + \frac{1}{2}at^{2}[/math]. Taking the car's initial position to be 0, and the edge of the cliff to be at 20 m, we thus have [math]20 = \frac{65000}{3600}t + \frac{10}{2}t^{2}[/math]. Ignoring the negative root, this gives us [math]t = \frac{8}{9} \approx 0.889\textnormal{ s}[/math], which makes sense, given that 65 km/hr is already a little over 18 m/s.

 

John is absolutely correct. To continue with John's analysis, let's calculate the speed of the car as it goes over the edge of the cliff:

 

[math]\frac{d}{dx}\left(5 t^2+\frac{65000}{3600}t\right)=10 t+\frac{65000}{3600}[/math]

 

When [math]t=8/9\,\text{s}[/math], the car has traversed the [math]20\,\text{m}[/math] and is at the edge of the cliff with a speed in the postive [math]x[/math] direction equal to:

 

[math]v=\left(10\,\frac{\text{m}}{\text{s}^2}\right)\left(\frac{8}{9}\,\text{s}\right)+\frac{65000}{3600}\,\frac{\text{m}}{\text{s}}=26.9444\,\frac{\text{m}}{\text{s}}[/math]

 

Ignoring friction and air resistance, the only force exerted upon the car after it goes over the cliff is gravity. So, let's calculate the time it takes for the car to hit the ground:

 

[math]y = -4.9\,\frac{\text{m}}{\text{s}^2}\,\,t^2+147.63\,\,\text{m}[/math]

 

Setting [math]y=0[/math] and solving for [math]t[/math] we get:

 

[math]0= -4.9\,\frac{\text{m}}{\text{s}^2}\,\,t^2+147.63\,\,\text{m}[/math]

 

[math]t=5.48895 \,\,\text{s}[/math]

 

As the car was falling, it was also moving in the positive [math]x[/math] direction with a speed of [math]26.9444\,\,\text{m/s}[/math]. Using this information along with the time it took for the car to hit the ground, we can calculate how far the car went in the positive [math]x[/math] direction:

 

[math]\left(26.9444\,\frac{\text{m}}{\text{s}}\right)\left(5.48895 \,\,\text{s}\right)=147.896\,\,\text{m}[/math]

 

The following image plots the path of the car:

Edited November 24, 201312 yr by Daedalus

Thanks guys.

You're welcome!