

Lesson 1: Describing Motion with WordsIntroduction to the Language of Kinematics 
Lesson 6: Kinematic Equations and ProblemSolvingSample Problems and SolutionsEarlier in Lesson 6, four kinematic equations were introduced and discussed. A useful problemsolving strategy was presented for use with these equations and two examples were given which illustrated the use of the strategy. Then, the application of the kinematic equations and the problemsolving strategy to freefall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented. These problems allow any student of physics to test their understanding of the use of the four kinematic equations to solve problems involving the onedimensional motion of objects. You are encouraged to read each problem and practice the use of the strategy in the solution of the problem. Then click the button to check the answer or use the link to view the solution.

Given:

Find: 
d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s^{2})*(32.8 s)^{2}
d = 1720 m
Given:

Find: 
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}
Given:

Find: v_{f} = ?? 
d = (0 m/s)*(2.6 s)+ 0.5*(9.8 m/s^{2})*(2.6 s)^{2}
d = 33 m
v_{f} = v_{i} + a*t
v_{f} = 0 + (9.8 m/s^{2})*(2.6 s)
v_{f} = 25.5 m/s ( indicates direction)
Given:

Find: a = ?? 
a = (46.1 m/s  18.5 m/s)/(2.47 s)
a = 11.2 m/s^{2}
d = v_{i}*t + 0.5*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 v_{f}^{2} = v_{i}^{2} + 2*a*d)
Given:

Find: 
1.40 m = (0 m/s)*(t)+ 0.5*(1.67 m/s^{2})*(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
Given:

Find: d = ?? 
a = (444 m/s  0 m/s)/(1.80 s)
a = 247 m/s^{2}
d = v_{i}*t + 0.5*a*t^{2}
d = (0 m/s)*(1.80 s)+ 0.5*(247 m/s^{2})*(1.80 s)^{2}
d = 0 m + 400 m
d = 400 m
(Note: the d can also be calculated using the equation v_{f}^{2} = v_{i}^{2} + 2*a*d)
Given:

Find: 
(7.10 m/s)^{2} = (0 m/s)^{2} + 2*(a)*(35.4 m)
50.4 m^{2}/s^{2} = (0 m/s)^{2} + (70.8 m)*a
(50.4 m^{2}/s^{2})/(70.8 m) = a
a = 0.712 m/s^{2}
Given:

Find: 
(65 m/s)^{2} = (0 m/s)^{2} + 2*(3 m/s^{2})*d
4225 m^{2}/s^{2} = (0 m/s)^{2} + (6 m/s^{2})*d
(4225 m^{2}/s^{2})/(6 m/s^{2}) = d
d = 704 m
Given:

Find: 
d = (22.4 m/s + 0 m/s)/2 *2.55 s
d = (11.2 m/s)*2.55 s
d = 28.6 m
Given:

Find: 
(0 m/s)^{2} = v_{i}^{2} + 2*(9.8 m/s^{2})*(2.62 m)
0 m^{2}/s^{2} = v_{i}^{2}  51.35 m^{2}/s^{2}
51.35 m^{2}/s^{2} = v_{i}^{2}
v_{i} = 7.17 m/s
Given:

Find: t = ?? 
(0 m/s)^{2} = v_{i}^{2} + 2*(9.8 m/s^{2})*(1.29 m)
0 m^{2}/s^{2} = v_{i}^{2}  25.28 m^{2}/s^{2}
25.28 m^{2}/s^{2} = v_{i}^{2}
v_{i} = 5.03 m/s
To find hang time, find the time to the peak and then double it.
v_{f} = v_{i} + a*t
0 m/s = 5.03 m/s + (9.8 m/s^{2})*t_{up}
5.03 m/s = (9.8 m/s^{2})*t_{up}
(5.03 m/s)/(9.8 m/s^{2}) = t_{up}
t_{up} = 0.513 s
hang time = 1.03 s
Given:

Find: 
(521 m/s)^{2} = (0 m/s)^{2} + 2*(a)*(0.840 m)
271441 m^{2}/s^{2} = (0 m/s)^{2} + (1.68 m)*a
(271441 m^{2}/s^{2})/(1.68 m) = a
a = 1.62*10^{5} m /s^{2}
Given:

Find: 
0 m/s = v_{i} + (9.8 m/s^{2})*(3.13 s)
0 m/s = v_{i}  30.6 m/s
v_{i} = 30.6 m/s
Now use: v_{f}^{2} = v_{i}^{2} + 2*a*d
(0 m/s)^{2} = (30.6 m/s)^{2} + 2*(9.8 m/s^{2})*(d)
0 m^{2}/s^{2} = (938 m/s) + (19.6 m/s^{2})*d
938 m/s = (19.6 m/s^{2})*d
(938 m/s)/(19.6 m/s^{2}) = d
d = 47.9 m
Given:

Find: 
370 m = (0 m/s)*(t)+ 0.5*(9.8 m/s^{2})*(t)^{2}
370 m = 0+ (4.9 m/s^{2})*(t)^{2}
(370 m)/(4.9 m/s^{2}) = t^{2}
75.5 s^{2} = t^{2}
t = 8.69 s
Given:

Find: 
(0 m/s)^{2} = (367 m/s)^{2} + 2*(a)*(0.0621 m)
0 m^{2}/s^{2} = (134689 m^{2}/s^{2}) + (0.1242 m)*a
134689 m^{2}/s^{2} = (0.1242 m)*a
(134689 m^{2}/s^{2})/(0.1242 m) = a
a = 1.08*10^{6} m /s^{2}
(The  sign indicates that the bullet slowed down.)
Given:

Find: 
d = (0 m/s)*(2.6 s)+ 0.5*(9.8 m/s^{2})*(3.41 s)^{2}
d = 0 m+ 0.5*(9.8 m/s^{2})*(11.63 s^{2})
d = 57.0 m
(NOTE: the  sign indicates direction)
Given:

Find: 
(0 m/s)^{2} = v_{i}^{2} + 2*(3.90 m/s^{2})*(290 m)
0 m^{2}/s^{2} = v_{i}^{2}  2262 m^{2}/s^{2}
2262 m^{2}/s^{2} = v_{i}^{2}
v_{i} = 47.6 m /s
Given:

Find: t = ?? 
(88.3 m/s)^{2} = (0 m/s)^{2} + 2*(a)*(1365 m)
7797 m^{2}/s^{2} = (0 m^{2}/s^{2}) + (2730 m)*a
7797 m^{2}/s^{2} = (2730 m)*a
(7797 m^{2}/s^{2})/(2730 m) = a
a = 2.86 m/s^{2}
v_{f} = v_{i} + a*t
88.3 m/s = 0 m/s + (2.86 m/s^{2})*t
(88.3 m/s)/(2.86 m/s^{2}) = t
t = 30. 8 s
Given:

Find: 
(112 m/s)^{2} = (0 m/s)^{2} + 2*(a)*(398 m)
12544 m^{2}/s^{2} = 0 m^{2}/s^{2} + (796 m)*a
12544 m^{2}/s^{2} = (796 m)*a
(12544 m^{2}/s^{2})/(796 m) = a
a = 15.8 m/s^{2}
Given:

Find: t = ?? 
v_{f}^{2} = v_{i}^{2} + 2*a*d
(0 m/s)^{2} = v_{i}^{2} + 2*(9.8 m/s^{2})*(91.5 m)
0 m^{2}/s^{2} = v_{i}^{2}  1793 m^{2}/s^{2}
1793 m^{2}/s^{2} = v_{i}^{2}
v_{i} = 42.3 m/s
Now convert from m/s to mi/hr:
v_{i} = 42.3 m/s * (2.23 mi/hr)/(1 m/s)
v_{i} = 94.4 mi/hr
Lesson 6: Kinematic Equations and ProblemSolving
 The Kinematic Equations
 Kinematic Equations and ProblemSolving
 Kinematic Equations and Free Fall
 Sample Problems and Solutions
 Kinematic Equations and Graphs
Other Resources: Physics Home Page  Multimedia Physics Studios  Shockwave Physics Studios 
Minds On Physics Internet Modules  The Review Session
© Tom Henderson
19962007