Mocha
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Post by Mocha on Feb 19, 2018 14:04:05 GMT
This puzzle is modified from one I read a few weeks ago. Hopefully it will make you think, and not just about math!
A two cars are driving down a straight road, parallel, in the same direction. The cars are identical. One car is travelling at 70 mi/h, the other is travelling 80 mi/h. Once the faster car overtakes the smaller car, they both simultaneously notice a fallen tree blocking the road. (ie. they are both the same distance from the tree) They both immediately begin to brake. The slower car barely stopped a hair's width away from the tree. However, logically, that means the faster car hit the tree. How fast did the faster car hit the tree?
Bonus Question 1: At what speed would the fast car have to be travelling for the impact speed to be 20 mi/h?
Bonus Question 2: The same scenario as the original, but instead of 70 mi/h, x mi/h, and instead of 80 mi/h, (x+10) mi/h.
All three problems are solveable, either with calculus alone, or with algebra + physics. There is no tricky wording, but if you don't think carefully, you'll derive the trick answer instead of the correct one. As a super bonus for those who can solve it both ways, prove the formula you learned in physics to be true using calculus!
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hamad
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Post by hamad on Mar 2, 2018 14:59:51 GMT
To me this seems very basic, but maybe I am missing something for the first car to stop it means vf=0 Normally breaking is constant deceleration, thus vf=v0+at 0=70-a*t a=70/t for the second car, we have the identical car, so it has same breaking deceleration, and the same time elapsed so same t we have vf=v0-at=80-70/t *t = 10 mi/h bonus again just change numbers so you need 90 90-70/t *t = 20mi/h for the second bonus a=x/t vf=(x+10)-x/t *t = 10mi/h
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R4DIC4L
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Post by R4DIC4L on Mar 10, 2018 6:37:57 GMT
The faster car, hits the obstacle at a speed of 10 miles/hour.
Bonus 1: The faster car should be going at a speed of 90 miles/hour at the point of it overtaking the slower car.
Bonus 2: The faster car will still have an impact at the speed of 10 miles/hour.
I'll provide a calculus based solution:
Let's assume that the time of overtake would be t=0, and the stopping acceleration for both cars are a constant called 'a' < 0 :
Acceleration-Time function for both cars: A1&2(t) = a
Integrating 'A1&2(t)' would give us a generic Velocity-Time function:
Int(A1&2(t)*dt) = Int(a*dt) = a*t + C => V1&2(t) = a*t + C
'C' however would be different values for Car #1 & #2:
V1(0) = 70 = A*(0) + C => C = 70 for V1(t) => V1(t) = a*t+70 V2(0) = 80 = A*(0) + C => C = 80 for V2(t) => V1(t) = a*t+80
Now we know that from a certain amount of time from t=0, V1(T) = 0 & V2(T) = ? : V1(T) = 0 = a*T + 70 => T = | 70/a | V2(| 70/a |) = a*| 70/a | + 80 = -70 + 80 = 10
=> Faster car will have a speed of 10 miles/hour at impact.
For the bonus ones you can repeat the same process by plugging and chugging different values into the V1(t) & V2(t) functions.
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Post by narenderkhola on Mar 25, 2018 8:28:57 GMT
According to equation of motion v = u + at
v= Final Speed = 0 u = Initial speed a = acceleration (Breaking here) t = time
For first car 0 = 70 + at t = -70/a
Speed of 2nd car at that time v = u+ at v = 80 + a(-70/a) v= 10 mi/hour
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