Maxima Chameleon
2 physics help with velocity and distance?
You are driving down the highway late one night at 20m/s when a deer steps onto the road 35m in front of you. Your reaction time before stepping on the brakes is 0.50s , and the maximum deceleration of your car is 10m/s^2 . How much distance is between you and the deer when you come to a stop?
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220m/s^2 for 20ms , then travels at constant speed for another 30ms. During this total time of 50ms , 1/20 of a second, how far does the tongue reach?
Answers:
(1) The distance between you and the deer is 5 m. Bambi barely lives.
(2) Tongue goes 0.176 m
How to find it and why it works
(1) Spooky and very stupid deer.
Given values in SI units:
Vi = 20 m/s
Vf = 0.0 m/s
d = 35 m
Tr = 0.50 s
a = -10 m/s^2 --> Note this must be negative, as acceleration is "acting against" your line of travel. if it isn't, your math won't work.
First calculate how far you travel before you hit the brakes. This is easy, 6th grade math. Distance, rate, time.
D = R * T
D = Vi * Tr
D = (20 m/s) * (0.50 s)
D = 10 m
Second, how far you went after you hit the brakes.
Dd = [ Vf^2 - Vi^2 ] / [ 2 * (-10 m/s^2) ]
Dd = [ (0.0 m/s)^2 - (20 m/s)^2 ] / (-20 m/s^2)
Dd = [ -400 m^2/s^2 ] / (-20 m/s^2)
Dd = 20 m
Total distance traveled
Dt = D + Dd
Dt = (10 m) + (20 m)
Dt = 30 m
Does he survive?
Ds = d - Dt
Ds = (35 m) - (30 m)
Ds = 5 m
So you come to a stop 5 meters before hitting him.
(1) Chameleons French kissing
Given values in SI units:
a = 220 m/s^2
t1 = 0.02 s
t2 = 0.03 s
Vi = 0.0 m/s
First calculate how far it travels during acceleration. To do this, initially find the final velocity of the tongue.
Vf = Vi + [a * t1]
Vf = (0.0 m/s) + [ (220 m/s^2) * (0.02 s) ]
Vf = 4.4 m/s
Then find how far it traveled as it accelerated.
d = [Vf^2 - Vi^2] / [2a]
d = [ (4.4 m/s)^2 - (0.0 m/s)^2 ] / [ 2 * (220 m/s^2) ]
d = [ 19.36 m^2/s^2 ] / [ 440 m/s^2 ]
d = 0.044 m
Now again, simple 6th grade math using the velocity of 4.4 m/s as your constant
D = R * t2
D = (4.4 m/s) * (0.03 s)
D = 0.132 m
Total distance covered
Dt = d + D
Dt = (0.044 m) + (0.132 m)
Dt = 0.176 m
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