Holt Physics Problem 2D - Hays High School

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Givens Solutions
4. vi = vavg = 518 km/h
vf = (0.600) vavg ∆t = 2.00 min
5. ∆t = 30.0 s
vi = 30.0 km/h
vf = 42.0 km/h
6. vf = 96 km/h
vi = 0 km/h
∆t = 3.07 s
7. ∆x = 290.0 m
∆t = 10.0 s
vf = 0 km/h = 0 m/s
8. ∆x = 5.7 × 10 3 km
∆t = 86 h
vf = vi + (0.10) vi 9. vi = 2.60 m/s
vf = 2.20 m/s
∆t = 9.00 min
Additional Practice 2D
1. vi = 186 km/h
vf = 0 km/h = 0 m/s
a =−1.5 m/s 2
2. vi =−15.0 m/s
vf = 0 m/s
a =+2.5 m/s 2
vi = 0 m/s
vf =+15.0 m/s
a =+2.5 m/s
vavg = 518 ⎯ � km
1 h
⎯��⎯⎯ ⎯
h 60 min��
1
3
0 m
⎯ = 8.63 × 10
1 km�
3 m/min
∆x = ⎯ 1
2 ⎯(vi + vf)∆t = ⎯ 1
2 ⎯[vavg + (0.600) vavg]∆t = ⎯ 1
2 ⎯(1.600)(8.63 × 10 3 m/min)(2.00 min)
∆x = 13.8 × 10 3 m = 13.8 km
2 ⎯(30.0
1h
km/h + 42.0 km/h) ⎯⎯ (30.0 s) �3600s� m 1 h
⎯�72.0 ⎯k ⎯��⎯⎯ (30.0 s)
h 3600s�
∆x = 3.00 × 10 −1 km = 3.00 × 10 2 m
∆x = ⎯ 1
2 ⎯(v i + v f)∆t = ⎯ 1
∆x = ⎯ 1
2
∆x = ⎯ 1
2 ⎯(v i + v f)∆t = ⎯ 1
2 ⎯(0
1h
km/h + 96 km/h) ⎯⎯ ⎯ �3600s�� 1
3
0 m
⎯ (3.07 s)
1 km�
∆x = ⎯ 1
2 ⎯� 96 × 103 ⎯ m
h ⎯ � (8.53 +×10−4 h) = 41 m
vi = ⎯ 2∆x
⎯ − vf = ⎯
∆t
(2)( 290.0
m)
⎯ − 0 m/s = 58.0 m/s = 209 km/h
10.
0 s
(Speed was in excess of 209 km/h.)
vf + vi = ⎯ 2∆x
⎯
∆t
vi (1.00 + 0.10) + vi = ⎯ 2∆x
⎯
∆t
vi = ⎯ (2)
3
( 5.
7 × 10
km)
⎯ = 63 km/h
( 2.
10)(
86
h)
∆x = ⎯ 1
2 ⎯(vi + vf)∆t = ⎯ 1
2 ⎯(2.60
60 s
m/s + 2.20 m/s)(9.00 min) ⎯ �min ∆x = 1.30 × 10 3 m = 1.30 km
⎯ � = ⎯ 1
∆t = ⎯ vf − vi ⎯ = =⎯
a
−
0 m/s − (186 km/h)� 51.7
m/
s
⎯ 2 = 34 s
−1.5
m/s
⎯ 1 h ⎯�� 3600s
⎯1
3
0 m ⎯� 1 km
⎯⎯⎯⎯
−1.5 m/s2 For stopping:
∆t1 = ⎯ vf − vi 0 m/s − (−15.
0 m/s)
⎯ =⎯⎯ a
2.5
m/
s2
= ⎯ 15.0
m/
s
⎯ 2 = 6.0 s
2.5
m/s
For moving forward:
∆t 2 = ⎯ vf − vi 15.0 m/s − 0.0 m/s 15.0
m/
s
⎯ = ⎯⎯ = ⎯⎯ 2 = 6.0 s
a 2.5 m/s2 2.5
m/s
∆t tot =∆t 1 +∆t 2 = 6.0 s + 6.0 s = 12.0 s
2 ⎯(4.80 m/s)(5.40 × 10 2 s)
Section Two — Problem Workbook Solutions II Ch. 2–5
II