Some object falls off a tower. A flying cat at the top of the tower flies down 3 sec later to catch the object. The cat's initial v = 500 cm/s and constant acceleration = 1400 cm/s^2. what is the time it takes for the flying cat to catch the object? how far does the object fall before it gets caught by the cat?
Given that the acceleration is = 1400 and Since the acceleration function a(t) is the rate of change of the velocity function v(t) then we can say that ∫a(t)dt= v(t). Therefore v(t) = 1400t+c. Since the cats initial velocity is 500 cm/s and he was 3 seconds late then we can say that v(3) = 500 or -3700 = c. So v(t) = 1400t-3700. The position function p(t) for the cat is then p(t) = 1400t^2 -3700t + c. If we make the position at the top of the tower 0 then p(3) = 0 and c = -1500 making p(t) = 1400t^2 -3700t - 1500 For the object, if we are assuming that the object is falling on earth, then the acceleration a(t) = 980 cm/s^2. v(t) is then v(t) = 980t+c. Assuming the initial velocity at time t = 0 is 0 then v(0) = 0 and c = 0 and v(t) = 980t. p(t) for the object is p(t) = 980t^2 + c. p(0) = 0 so c = 0 and p(t)= 980t^2. set them equal and 980t^2 = 1400t^2 -3700t - 1500. solve for t by subtracting 980t^2 from both sides then 0 = 420t^2 -3700t - 1500 or 0 = 42t^2 -370t - 150. use the quadratic formula and t = 5/42 (37+sqrt(1621)).
It's a trick question. Cats can't fly. ;)
@eighthourlunch no further comments?
Ha! Sorry, no. I shouldn't have answered that much, I'm supposed to be helping with dinner.
thanks But shouldn't p(t) = 490t^2? and p(t) for superman 700t^2 +......
Ah @Algebraic! found a mistake in my calculations. The position equation for the cat should be 700t^2 -3700t - 1500 and the position equation for the object should be 490t^2 making t = 5/21 (37+sqrt(1495))
@FishingWithTNT yes sorry about that.
oops messed up again, forgot to solve for c with the new equations. I think you get the gist of what I was trying to do though?
yeah I got it! thanks
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