Holt Physics Problem 2D - Hays High School

Copyright © by Holt, Rinehart and Winston. All rights reserved.
Givens Solutions
8. vi = 0 m/s
vf = 3.06 m/s
a = 0.800 m/s 2
∆t 2 = 5.00 s
9. vf = 3.50 × 10 2 km/h
vi = 0 km/h = 0 m/s
a = 4.00 m/s 2
10. vi = 24.0 m/s
a =−0.850 m/s 2
∆t = 28.0 s
11. a =+2.67 m/s 2
∆t = 15.0 s
∆x =+6.00 × 10 2 m
12. a = 7.20 m/s 2
∆t = 25.0 s
vf = 3.00 × 10 2 ms
13. vi = 0 m/s
∆x = 1.00 × 10 2 m
∆t = 12.11 s
14. vi = 3.00 m/s
∆x = 1.00 × 10 2 m
∆t = 12.11 s
15. vf = 30.0 m/s
vi = 18.0 m/s
∆t = 8.0 s
∆t1 = ⎯ vf − vi 3.06
m/
s − 0
⎯ =⎯
a 0.
800
m/
s 2
m/s
⎯ = 3.82
∆x 1 = v i∆t 1 + ⎯ 1
2 ⎯a∆t 1 2 = (0 m/s) (3.82 s) + ⎯ 1
2 ⎯(0.800 m/s 2 ) (3.82 s) 2 = 5.84 m
∆x 2 = v f∆t 2 = (3.06 m/s)(5.00 s) = 15.3 m
∆x tot =∆x 1 +∆x 2 = 5.84 m + 15.3 m = 21.1 m
∆t = ⎯ (v (3.50 × 10
f − vi) ⎯ = = 24.3 s
a
2 km/h − 0 km/h)�⎯ 1 h ⎯�� 3600
s
⎯1
3
0 m ⎯� 1 km
⎯⎯⎯⎯⎯
(4.00 m/s2 )
∆x = v i∆t + ⎯ 1
2 ⎯a∆t 2 = (0 m/s)(24.3 s) + ⎯ 1
2 ⎯(4.00 m/s 2 )(24.3 s) 2
∆x = 1.18 × 10 3 m = 1.18 km
v f = v i + a∆t = 24.0 m/s + (− 0.850 m/s 2 )(28.0 s) = 24.0 m/s − 23.8 m/s = +0.2 m/s
vi ∆t =∆x− ⎯ 1
⎯a∆t
2
2
vi = ⎯ ∆x
∆t
⎯ − ⎯ 1
v i = v f − a∆t
2 ⎯a∆t = ⎯ 6.00
2
× 10 m
15.0
s
⎯ − ⎯ 1
2 ⎯(2.67 m/s 2 )(15.0 s) = 40.0 m/s − 20.0 m/s = +20.0 m/s
v i =(3.00 × 10 2 m/s) − (7.20 m/s 2 )(25.0 s) = (3.00 × 10 2 m/s) −(1.80 × 10 2 m/s)
v i = 1.20 × 10 2 m/s
∆x = vi∆t + ⎯ 1
⎯a∆t
2
2
Because vi = 0 m/s,
a = ⎯ 2∆
∆t
2
x
⎯ =⎯ (2)(
2
1.00
× 10
m)
⎯
( 12.11
s) 2 = 1.36 m/s 2
a = ⎯ 2(∆x −
∆t
2
v
⎯
i∆t) (2)[1.00 × 10
=
2 m − (3.00 m/s)(12.11 s)]
⎯⎯⎯⎯
(12.11 s) 2
a =
(2)(1.00 × 10 2 m − 36.3 m)
⎯⎯⎯
(12.11 s) 2
a = ⎯ ( 2)
( 64
m
( 12.
11
s) 2
)
⎯ = 0.87 m/s 2
a = ⎯ vf − vi ⎯ = =⎯
∆t
12.0
m/s
⎯ = 1.5 m/s
8.0
s
2
30.0 m/s − 18.0 m/s
⎯⎯
8.0 s
Section Two — Problem Workbook Solutions II Ch. 2–7
II