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