Summary of Equations

D = vt

Two objects (A and B) with same (constant) velocity:
D_B/D_A = t_B/t_A

Definition:
acceleration = (v_final - v_initial)/t

Object starting from rest, with constant acceleration:
v = at [v_final is now written as "v".]
D = (1/2)at²
v_average = (1/2)v
D = v_averaget

Two objects (A and B) starting from rest with same acceleration:
D_B/D_A = (t_B/t_A)²

Problem 1

D = vt = 4.5 X 40 = 180 m.

Problem 2

D = vt = 4.5 X 120 = 540 m.

Problem 3

D_B/D_A = t_B/t_A
D_B/100 = 120/40 = 3
D_B = 100 X 3 = 300 m.

Problem 4

D_B/D_A = t_B/t_A
D_B/50 = 38/15 = 2.53
D_B = 50 X 2.53 = 127 m.
[Note that you don't have to change units (minutes to seconds, or kilometers to meters) as long as you use the same units for both objects.]

Problem 5

a = v_final - v_initial)/t
a = (315 - 185)/6.5 = 130/6.5 = 20 m/s²

Problem 6

v = at = 5 X 10 = 50 m/s.

Problem 7

D = (1/2)at² = (1/2)(20)(30)² = (1/2)(20)(900) = 9,000 m.

One minute = 60 seconds.
D = (1/2)at² = (1/2)(20)(60)² = (1/2)(20)(3600) = 36,000 m.

Problem 8

Call the airplanes A, B, and C:
t_A = 15 s.
t_B = 45 s.
t_C = 49 s.
D_A = 120 m.

t_B is three times t_A. Therefore D_B should be nine times D_A, since 3² = 9. Hence D_B = 9 X 120 = 1,080 m.

For D_C you can't do the calculation in your head, but use:
D_C/D_A = (t_C/t_A)².
D_C/120 = (49/15)² = 3.27² = 10.7,
D_C = 120 x 10.7 = 1,284 m.
[You'd get a slightly different answer if you did not round off the 3.27 and the 10.7 as I did above. Note also that you could do this problem using D_C/D_B instead of D_C/D_A, and the answer would come out the same.]

Problem 9

v = at = 18 X 6 = 108 m/s. This is the final speed.
v_average = (1/2)v = 54 m/s.

Problem 10

Let "A" represent the first 10 seconds, "B" the first 100 seconds, and "C" the first 125 seconds.
Since t_B is 10 times t_A, we must have D_B equal to 100 times D_A (10² = 100). So D_B = 100 X 80 = 8,000 m.
For case C, D_C/D_A = (t_C/t_A)²,
D_C/80 = (125/10)² = 12.5² = 156.25,
D_C = 80 X 156.25 = 12,500 m.

Problem 11

a = (v_final - v_initial)/t = (0.064 - 0)/4 = 0.016 m/s².
v_average = (1/2)v_final = (1/2)(0.064) = 0.032 m/s.

D = (1/2)at² = (1/2)(0.016)(4)² = (1/2)(0.016)(16) = 0.128 m.
Alternately, D = v_averaget = 0.032 X 4 = 0.128 m.

Problem 12

Horizontal motion is at constant velocity, v = 4.2 m/s. There is no acceleration, so distance = vt = 4.2 X 0.8 = 3.36 m.

Problem 13

Vertical motion has an acceleration, g = 9.8 m/s². This is independent of the horizontal motion. Vertical distance is given by D = (1/2)gt², or

D = (1/2) X (9.8) X (1.3)² = 4.9 X 1.69 = 8.28 m.