1. Extra Advanced Kinematic Math Problems

2. 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 = m

3. 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

4. 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 = 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 = m

5. 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: a = ?? 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 m
X = m
(Note: X can also be calculated using the equation Vf2 =Vi2+ 2×a×X)

6. -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 = m
a = 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+ ( m/s2) ×(t)2
(-1.40 m)/( m/s2) = t2
1.68 s2 = t2
t = 1.29 s

7. 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: a = ?? 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= 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 m
X = 400 m
(Note: X can also be calculated using the equation Vf2 =Vi2 + 2×a×X)

8. (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 = m/s2

9. (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

10. 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

11. (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 = vi m2/s2
51.35 m2/s2 = vi2
vi = 7.17 m/s