Pro Tip
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DISCLAIMER: I'm doing this problem the “ugly math” kind of way, there are easier ways.
PROTIP: always talk to the villagers twice. It takes the flag 0.1534 seconds to pass through the first gate. The floating thing traveled 20 cm before the light turned on, thus with the third kinematic equation (v1t omitted since initial velocity is zero)
x=1/2*at^2 1/2*at^2-x=0 (Quadratic formula) Tb=[(-4*1/2*a*-20)^(1/2)]/a=[(40a)^(1/2)]/a (Beginning of light) Te=[(-4*1/2*a*-30)^(1/2)]/a=[(60a)^(1/2)]/a (End of light) Te-Tb=0.1534, or [(60a)^(1/2)]/a-[(40a)^(1/2)]/a=0.1534
Eventually, a should be 85.85959829 m/s^2, left, I solved it by graphing both sides of the equation in the calculator, and finding the intercept (TI-83+). We use the third kinematic equation, again. x=1/2*at^2 180-100=80=1/2(85.8595...)t^2, and so t=1.365103... (Beginning of 2nd L) 190-100=90=1/2(85.8595...)t^2, and so t=1.4479019... (End of 2nd L) Thus the time the second light flashes is about 0.0828 seconds.
PROTIP: always talk to the villagers twice. It takes the flag 0.1534 seconds to pass through the first gate. The floating thing traveled 20 cm before the light turned on, thus with the third kinematic equation (v1t omitted since initial velocity is zero)
x=1/2*at^2 1/2*at^2-x=0 (Quadratic formula) Tb=[(-4*1/2*a*-20)^(1/2)]/a=[(40a)^(1/2)]/a (Beginning of light) Te=[(-4*1/2*a*-30)^(1/2)]/a=[(60a)^(1/2)]/a (End of light) Te-Tb=0.1534, or [(60a)^(1/2)]/a-[(40a)^(1/2)]/a=0.1534
Eventually, a should be 85.85959829 m/s^2, left, I solved it by graphing both sides of the equation in the calculator, and finding the intercept (TI-83+). We use the third kinematic equation, again. x=1/2*at^2 180-100=80=1/2(85.8595...)t^2, and so t=1.365103... (Beginning of 2nd L) 190-100=90=1/2(85.8595...)t^2, and so t=1.4479019... (End of 2nd L) Thus the time the second light flashes is about 0.0828 seconds.
