# Physics 1303 - Summer 1998

# Homework Assignment #2

**Due:** Monday 15 June 1998

### READING

Chapters 2 and 4.
### QUESTIONS

Chapter 2 - 1, 2, 4, 5, 6, 8, 10, 13.

Chapter 4 - 10, 12, 14, 15, 16, 18.
### PROBLEMS

Chapter 2 - 2, 3, 6, 19, 21, 30, 50, 58.

Chapter 4 - 3, 13, 28, 39, 47.
### ANSWERS

These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to
copy these and turn them in as homework. You must show your work.
**Q 2-1)** They are equal for the case of CONSTANT velocity.

**Q 2-2)** No, the instantaneous velocity can be zero during the interval.

**Q 2-4)** Yes.

**Q 2-5)** Yes.

**Q 2-6)** Yes.

**Q 2-8)** They are the same.

**Q 2-10)** The displacement vector is zero as well.

**Q 2-13)** 9.8 m/s each second.

**Q 4-10)** Yes, the point directly under the release point.

**Q 4-12)** No.

**Q 4-14)** a) No. b) No. In each case, the magnitude is constant,
but the direction of the vector changes.

**Q 4-15)** Yes, 9.8 m/s^{2} vertical, 0 m/s^{2} horizontal

**Q 4-16)** 60^{o}

**Q 4-17)** Moon, moon.

**Q 4-18)** b) acceleration, c) horizontal component of velocity

**P 2-2)**
- 180 km North
- 63 km/h North

**P 2-3)**
- 5 m/s
- 1.25 m/s
- -2.5 m/s
- -3.3 m/s
- 0 m/s

**P 2-6)**
- 50 m/s
- 41 m/s

**P 2-19)**
- 20.0 m/s, 5.00 m/s
- 262 m

**P 2-21)**
[The function should be x(t) = 2.0 + 3.0 t - 1.0 t^{2}]
- 2 m
- -3 m/s
- -2 m/s
^{2}

**P 2-30)**
- -2.25 x 10
^{-2} m/s^{2}
- 133 s
- 1.5 m/s

**P 2-50)**
- 98 m/s upward
- 490 m

**P 2-58)** 14.0 m

**P 4-3)**
- 4.87 km at 209
^{o} from East
- 23.3 m/s
- 13.5 m/s at 209
^{o} from East

**P 4-13)**
- 2.67 s
- 29.9 m/s horizontal
- (29.9 m/s, -26.2 m/s)

**P 4-28)**
- 6 rev/s gives the larger linear speed
- 1.52 x 10
^{3} m/s^{2}
- 1.28 x 10
^{3} m/s^{2}

**P 4-39)**
- 30.8 m/s
^{2} downward
- 70.4 m/s
^{2} upward

**P 4-47)** 153 km/h at 11.3^{o} North of West

Please report any corrections to
Professor Scalise.

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