Kinematic Equation Practice

Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? (You will need to use TWO equations. Which ones, and in which order?)

Given: a = -9.8 m t = 2.6 s vi = 0 m/s Find: d = ?? vf = ?? Equation: d = (vi)(t) + (1/2)(a)(t^2) d = (0 m/s)(2.60 s) + 0.5(-9.8 m/s2)(2.60 s)^2 d = -33.1 m (- indicates direction) vf = (vi) + (a)(t) vf = 0 + (-9.8 m/s2)(2.60 s) vf = -25.5 m/s (- indicates direction)

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: d = 110 m t = 5.21 s vi = 0 m/s Find: a = ?? Equation: d = (vi)(t) + (1/2)(a)(t^2) 110 m = (0 m/s)(5.21 s) + 0.5(a)(5.21 s)^2 110 m = (13.57 s^2)(a) a = (110 m)/(13.57 s^2) a = 8.10 m/ s^2

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/s^2. Determine the time for the feather to fall to the surface of the moon.

Given: vi = 0 m/s d = -1.40 m a = -1.67 m/s^2 Find: t = ?? Equation: d = (vi)(t) + (1/2)(a)(t^2) -1.40 m = (0 m/s)(t) + 0.5(-1.67 m/s2)(t)^2 -1.40 m = 0 + (-0.835 m/s^2)(t)^2 (-1.40 m)/(-0.835 m/s^2) = t^2 1.68 s^2 = t^2 t = 1.29 s

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 d = 35.4 m Find: a = ?? (vf)^2= (vi)^2+ 2(a)(d) (7.10 m/s)^2 = (0 m/s)^2 + 2(a)(35.4 m) 50.4 m2/s^2 = (0 m/s)^2 + (70.8 m)(a) (50.4 m2/s^2)/(70.8 m) = a a = 0.712 m/s^2

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: d = ?? a = ?? a = (Delta v)/(t) (This is the definition of acceleration, not necessarily a kinematic equation) a = (46.1 m/s - 18.5 m/s)/(2.47 s) a = 11.2 m/s2 d = (vi)(t) + (1/2)(a)(t^2) d = (18.5 m/s)(2.47 s) + 0.5(11.2 m/s^2)(2.47 s)^2 d = 45.7 m + 34.1 m d = 79.8 m (Note: the d can also be calculated using the equation vf^2 = vi^2 + 2ad)

An airplane accelerates down a runway at 3.20 m/s^2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

Given: a = +3.2 m/s^2 t = 32.8 s vi = 0 m/s (assumed) Find: d = ?? Equation: d = (vi)(t) + (1/2)(a)(t^2) d = (0 m/s)(32.8 s)+ 0.5(3.20 m/s2)(32.8 s)^2 d = 1720 m