Extra Advanced Kinematic Math Problems

Equation: X = Vi*t + 0.5*a*t2 1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until it finally lifts off the ground. Determine the distance traveled before takeoff. Given: a = 3.2 m/s2 t = 32.8 s Vi = 0 m/s Find: X = ?? Equation: X = Vi*t + 0.5*a*t2 X =(0 m/s)×(32.8 s)+ 0.5×(3.20 m/s2)×(32.8 s)2 X = 1721.3 m

110 m = (0 m/s)×(5.21 s)+ 0.5×(a)×(5.21 s)2 110 m = (13.57 s2)×a 2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car. Given: X = 110 m t = 5.21 s Vi = 0 m/s Find: a = ?? Equation: X = Vi×t + 0.5×a×t2 110 m = (0 m/s)×(5.21 s)+ 0.5×(a)×(5.21 s)2 110 m = (13.57 s2)×a a = (110 m)/(13.57 s2) a = 8.10 m/ s2

vf = -25.48 m/s (- indicates direction) 2. X = Vi×t + 0.5×a×t2 3. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall? Given: a = -9.8 m/s2 t = 2.6 s vi = 0 m/s Find: 1. vf = ?? 2. X = ?? 1. vf = vi + a×t vf = 0 + (-9.8 m/s2)×(2.6 s) vf = -25.48 m/s (- indicates direction) 2. X = Vi×t + 0.5×a×t2 X = (0 m/s)×(2.6 s)+ 0.5×(-9.8 m/s2)×(2.6 s)2 X = 33.12 m

4. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. Given: Vi = 18.5 m/s Vf = 46.1 m/s t = 2.47 s Find: 1. a = ?? 2. X= ?? 1. Vf= Vi + a×t 46.1m/s = 18.5 m/s+ a×(2.47 s) 46.1m/s -18.5m/s= a×(2.47) (27.6m/s)/2.47s = a×(2.47 s)/(2.47 s) A=11.17 m/s2 2. X = Vi×t + 0.5×a×t2 X = (18.5 m/s)×(2.47 s)+ 0.5×(11.17 m/s2)×(2.47 s)2 X = 45.7 m + 34.1 m X = 79.78 m (Note: X can also be calculated using the equation Vf2 =Vi2+ 2×a×X)

-1.40 m = (0 m/s)×(t)+ 0.5× (-1.67 m/s2)×(t)2 5. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. Given: Vi = 0 m/s X = -1.40 m a = -1.67 m/s2 Find: t = ?? X = Vi×t + 0.5×a×t2 -1.40 m = (0 m/s)×(t)+ 0.5× (-1.67 m/s2)×(t)2 -1.40 m = 0+ (-0.835 m/s2) ×(t)2 (-1.40 m)/(-0.835 m/s2) = t2 1.68 s2 = t2 t = 1.29 s

6. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels? Vi = 0 m/s Vf = 444 m/s t = 1.80 s Given: Find: 1. a = ?? 2. X = ?? 1. Vf = Vi + a×t 444m/s = 0 m/s+ a(1.8 s) (44m/s)/(1.8s)= [a(1.8 s)]/1.8s a= 246.67 m/s2 2. X = Vi× t + 0.5×a×t2 X = (0 m/s) ×(1.80 s)+ 0.5× (247 m/s2) ×(1.80 s)2 X = 0 m + 400 m X = 400 m (Note: X can also be calculated using the equation Vf2 =Vi2 + 2×a×X)

(7.10 m/s)2 = (0 m/s)2 + 2×(a)×(35.4 m) 7. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike. Given: Vi = 0 m/s Vf = 7.10 m/s X= 35.4 m Find: a = ?? Vf2 = Vi2 + 2×a×X (7.10 m/s)2 = (0 m/s)2 + 2×(a)×(35.4 m) 50.4 m2/s2 = (0 m/s)2 + (70.8 m)×a (50.4 m2/s2)/(70.8 m) = a a = 0.712 m/s2

(65 m/s)2 = (0 m/s)2 + 2×(3 m/s2)×X 4225 m2/s2 = (0 m/s)2 + (6 m/s2)×X Find: d = ?? An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway? Given: Vi = 0 m/s Vf = 65 m/s a = 3 m/s2 Find: X = ?? vf2 = vi2 + 2×a×X (65 m/s)2 = (0 m/s)2 + 2×(3 m/s2)×X 4225 m2/s2 = (0 m/s)2 + (6 m/s2)×X (4225 m2/s2)/(6 m/s2) = X X = 704 m

X = ½ (Vi + Vf)×t X = ½(22.4 m/s + 0 m/s)×2.55 s 9. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration). Given: Vi = 22.4 m/s Vf = 0 m/s t = 2.55 s Find: X = ?? X = ½ (Vi + Vf)×t X = ½(22.4 m/s + 0 m/s)×2.55 s X = (11.2 m/s) ×2.55 s X= 28.6 m

(0 m/s)2 = vi2 + 2× (-9.8 m/s2) ×(2.62 m) 0 m2/s2 = vi2 - 51.35 m2/s2 10. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo. Given: a = -9.8 m/s2 Vf = 0 m/s X = 2.62 m Find: Vi = ?? vf2 = vi2 + 2×a×X (0 m/s)2 = vi2 + 2× (-9.8 m/s2) ×(2.62 m) 0 m2/s2 = vi2 - 51.35 m2/s2 51.35 m2/s2 = vi2 vi = 7.17 m/s