PHYS207 Honors

Answers For a Few Exam Problems

Exam 1

1996:
1.
(a) Any of the following, and perhaps others:

1. Explain how to use units or explain the physical meaning of each term in the equation for the distance; then differentiate to verify that a(t) = constant acceleration.

2. Explain how to use units or explain the physical meaning of both the distance and the velocity equation.

3. Set a(t) = a = constant and integrate twice.

(b)

v_x = maximum roughly where x(t) crosses x=0 going up
v_x = minimum roughly where x(t) crosses x=0 going down
v_x = 0 where x(t) is a maximum or minimum.

2.
a_c = 4 pi² R cos(latitude) / (1 day)² = 0.03 m / sec².

3.
a = [v² - v₀²] / [2 d] = - 2.4 x 10¹⁷ m / sec²
where v and v₀ are the final and initial velocities, resp., and d is the thickness of the paper.
t = 2 d / [v + v₀ ] = 3.3 x 10 ^-11 sec

4. Note that the height of the slope at any given x is y = - x tan(phi)

x_max = 2 v₀² cos²(theta) [tan (theta) + tan (phi)] / g
The angle for maximum range is theta_max = pi/4 - phi/2

1995:
2.
The time of flight is d/v₀ = 0.473 sec (where d is the distance to home plate).
In this length of time the ball falls a distance y = g d² / [2 v₀² ] = - 1.1 m.

3.
r = cos(wt) 1_x + sin(wt) 1y
v = w [ - sin(wt)1_x + cos(wt)1_y ]
a = w² [- cos(wt) 1_x - sin(wt) 1y]
   = - w² r
j = w³ [sin(wt)1_x - cos(wt)1_y]
   = - w² v

4. If t_d is the delay time (5 sec) and v₀ is the velocity of the passing car
t = (v₀/a) + t_d + sqrt[(v₀/a)² + 2 v₀ t_d / a]
   = 49.5 sec

d = v₀ t = 1580 m