Star Trek Car Chase Scene Basic Information

1965 corvette Sting Ray

Dimensions- L=175in or 14.58ft, W= 69.6in or 5.8ft

Weight= 3230 lbs or 1465.1 kg What Are We Looking For?

Will Lil' Captain James T. Kirk Survive? What Information Do we Have

We will asume the speed remains around 80 mph or 35.76m/s

From the still shot we were able to figure out the distance from the cliff to the car when he puts on the breaks With the distance from the car to the cliff 1.9 car lengths away we were able to determine how far away the car actually was by: 175 x 1.9= 332.9/12= 27.7 ft

27.7ft x .3048m= 8.44m away What We Need to Find In order to determine if Lil' Kirk will survive there are three factors we will need to determine:

1. After the brakes are set, what is the car's deceleration rate as it slides toward the cliff

2. How fast is Kirk moving after he jumps from the car

3. How quickly will Kirk need to decelerate in order to prevent himself from falling to his death 1. How fast is the car moving toward the cliff

Fk=uFn

u= .7 (for asphalt)

Fn= mg; m=1465.1 kg, g= 9.8

Fk= (.7)(1465.1)(9.8)= 10050.586

Ef=ma so Fk=ma

10050.586= ma

a=10050.586/1465.1= -6.86m/s^2 In order to find the speed of the car when it falls over the edge:

Vf= Vi + at

Vi= car's speed or 35.76 m/s

a= -6.86m/s^2

t= 3.1 s (a rough estimate)

Vf= 35.76m/s + (-6.86m/s^2)(3.1 s)

Vf= 35.76m/s - 21.26

Vf= 14.49m/s or 32.42 mph; speed of the car when it falls 2. How fast is Kirk moving after he jumps

Lets assume that Kirk jumps out of the car at 3 m/s

Speed of the car = Vf: 14.49m/s or 32.42 mph

So Vf- 3 m/s= Kirk's speed as he is moving towards the cliff

(14.49)- 3 m/s= 11.49m/s is the speed at which Kirk is traveling toward the cliff from an assumed distance of 3 meters away which is the point where he left the car

So once Kirk leaves the car he is still moving towards the cliff at 11.49m/s or 25.7 mph 3. How fast will Kirk need to slow down to live

After jumping from the car Kirk is now traveling at a speed of 11.49m/s or 25.7 mph

To solve this we just use: Vf^2= Vi^2 + 2a(^x); (^x)= change in x

So; 0= 11.49^2 + 2a(3m)

a= -22.00m/s

Then to see what kind of coefficient of friction he will need to come to a stop in 3m we use; Fk=ma, where m is a guess of 120lbs or 54.5 kg for Kirk's weight and a= -22.00m/s

Fk= -1199

Fk/Fn= u

So -1199/(54.5)(9.8)= -2.24= u= How much friction Kirk will need to come to a stop What Does This Mean For Kirk?

He Would Not Survive Based on the information given and taken from the clip, young Kirk, still traveling at 11.49m/s or 25.7 mph 3m away from the cliff would not be able to slow down fast enough to come to a stop at the very edge of the cliff.

In actuallity Kirk would slide right off the cliff along with the corvette and the movie would have ended right there.

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# Star Trek Physics Project

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