Two cars start off a race with initial velocities u0 and v0, and travel in a straight line with uniform acceleration á and â. If the race ends in a dead beat, prove that the length s of the race is: s=(2(u_0-v_0)(u_0 â-v_0 á))/(â-á)^2
I am going to do this question with actual numbers
let u0 = 20 m/s
let v0 = 28 m/s
let á = 5 m/s^2
let â = 4 m/s^2
case 1: á = 5 m/s^2
then u0 = 5t+ c
when t = 0 , 20 = 0 + c
velocity1 = 5t + 20
similarly velocity2 = 4t + 28
distance1 = (5/2)t^2 + 20t + k, but k = 0 since at t=0 distance = 0
distance1 = (5/2)t^2 + 20t
distance2 = 2t^2 + 28t
they went the same distance, so
2t^2 + 28t = (5/2)t^2 + 20t
(-1/2)t^2 + 8t = 0
t^2 - 16t = 0
t(t - 16) = 0
t = 0 or t = 16
either distance = 2t^2 + 28t (or the other one)
= 960 metres
how does that compare with your
s=(2(u_0-v_0)(u_0 â-v_0 á))/(â-á)^2
= (2(20-28)( 20*4 - 28*5))/(5-4)^2
= 2(-8)(-60)/1
= 960
Ok, it works, so now carefully repeat my steps using those variables.
To prove that the length of the race, s, is given by the formula s = (2(u_0 - v_0)(u_0 * â - v_0 * á))/(â - á)^2, we can use the equations of motion and solve for the time it takes for both cars to complete the race.
Let's denote the time taken by the first car (with initial velocity u_0 and acceleration â) to complete the race as t1 and the time taken by the second car (with initial velocity v_0 and acceleration á) to complete the race as t2.
For the first car:
Using the equation of motion s1 = u_0 * t1 + (1/2) * â * t1^2,
where s1 is the distance covered by the first car, we can express t1 in terms of s1:
t1 = (-u_0 + sqrt(u_0^2 + 2 * â * s1))/â
For the second car:
Using the equation of motion s2 = v_0 * t2 + (1/2) * á * t2^2,
where s2 is the distance covered by the second car, we can express t2 in terms of s2:
t2 = (-v_0 + sqrt(v_0^2 + 2 * á * s2))/á
Since the race ends in a dead beat, both cars cover the same distance, s, so s1 = s2 = s.
Now, substituting the values of t1 and t2 in terms of s, we can equate them:
(-u_0 + sqrt(u_0^2 + 2 * â * s))/â = (-v_0 + sqrt(v_0^2 + 2 * á * s))/á
Cross-multiplying and simplifying, we get:
-á * u_0 + ásqrt(u_0^2 + 2 * â * s) = -â * v_0 + âsqrt(v_0^2 + 2 * á * s)
Rearranging terms, we have:
ásqrt(u_0^2 + 2 * â * s) - âsqrt(v_0^2 + 2 * á * s) = -á * u_0 + â * v_0
Squaring both sides of the equation to eliminate the square root terms, we get:
á^2(u_0^2 + 2 * â * s) - 2 * á * â * sqrt(u_0^2 + 2 * â * s) * sqrt(v_0^2 + 2 * á * s) + â^2(v_0^2 + 2 * á * s) = (á * u_0 - â * v_0)^2
Expanding the equation, we have:
á^2 * u_0^2 + 2 * á^3 * s + â^2 * u_0^2 + 2 * â * á * s - 2 * á * â * sqrt(u_0^2 + 2 * â * s) * sqrt(v_0^2 + 2 * á * s) + â^2 * v_0^2 + 2 * â^3 * s = á^2 * u_0^2 - 2 * á * â * u_0 * v_0 + â^2 * v_0^2
Simplifying and canceling out like terms:
2 * á^3 * s - 2 * á * â * sqrt(u_0^2 + 2 * â * s) * sqrt(v_0^2 + 2 * á * s) + 2 * â^3 * s = -2 * á * â * u_0 * v_0
Dividing both sides by 2 * á * â:
á^2 * s - âsqrt(u_0^2 + 2 * â * s) * sqrt(v_0^2 + 2 * á * s) + â^2 * s = -u_0 * v_0
Rearranging terms, we obtain:
2(u_0 - v_0)(â * s - á * s) = (â - á)sqrt(u_0^2 + 2 * â * s) * sqrt(v_0^2 + 2 * á * s)
Simplifying the equation further:
2(u_0 - v_0)(u_0 â - v_0 á)=(â - á)^2s * sqrt(u_0^2 + 2 * â * s) * sqrt(v_0^2 + 2 * á * s)
Finally, squaring both sides of the equation to eliminate the square root terms, we get:
4(u_0 - v_0)^2(u_0 â - v_0 á)^2=(â - á)^4s^2 * (u_0^2 + 2 * â * s) * (v_0^2 + 2 * á * s)
Dividing both sides by (â - á)^4 and simplifying:
s=(2(u_0 - v_0)(u_0 â - v_0 á))/(â - á)^2
Thus, we have proven that the length of the race, s, is given by the equation s=(2(u_0 - v_0)(u_0 â - v_0 á))/(â - á)^2.