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