PHYS 1111 Chapter 2 Problems

Solutions

1

 

a) Assume that down the pool is considered the positive direction. Since the displacement covered is +50.0 m and the time required is 20.0 s, the average velocity is +50.0/20.0 = +2.50 m/s.
b) The displacement is -50.0 m (opposite to the direction in part a) and the time is 22.0 s, so the average velocity is -50.0/22.0 = -2.27 m/s.
c) The total displacement is zero so the average veloctiy for the entire round trip is zero.

d) +2 m/s

2

 

The following information is given:
v_i=0, v_f=10.97 x 10³m/s, and x=220 m.
Using equation 5
2ax=v_f² - v_i² we get
a= 2.74 x 10⁵ m/s/s. This is approximately 27900 time the acceleration due to gravity!!

3

 

Given v_i = +20 m/s, v_f = +30 m/s, and
x = +200 m, use equation 5 to find:
a) a = +1.25 m/s/s
b) Using equation 3 we find t = (v_f - v_i)/a, or
t = 8.00 s.

4

 

Given v_i = 0, v_f =120 km/hr, and x = 240 m, we wish to find the acceleration and the time required.
a) Converting 120 km/hr to m/s, we get 1.20 x 10⁵m/hr x 1hr/3600s = 33.3 m/s. Using equation 5, we find a = 2.31 m/s/s.
b) Equation 3 can now be used to find a time t = 14.4 s.

5

 

(a) If we define up as the positive direction, then the ball has an initial velocity of v_o = 25 m/s, an initial displacement of y_o = 0, and a constant acceleration of -9.81 m/s². When the ball has reached its highest point, we know that the velocity of the ball is 0. Therefore, using equation 2.10 (v² = v_o² + 2 a (y_f - y_o)), we get

0 = (25 m/s)² + 2(-9.81 m/s²)y_f.

Solving this for y_f, we get

y_f = (625 m²/s²)/(19.6 m/s²) = 31.9 m.

(b) To find the length of time before the ball reaches its highest point, we use equation 2.6 to get

0 = 25 m/s + (-9.8 m/s²)t

Solving for t gives

t = (25 m/s)/(9.81 m/s²) = 2.55 s.

(c) We now work the problem in reverse. Starting at the highest point, v_o = 0, y_o = 31.9 m, and y_f = 0 m. Using equation 2.9, we get

0 m - 31.9 m = (0 m/s)t + .5(-9.81 m/s²)t²
-31.9 m = (-4.91 m/s²)t²
t² = (-31.9 m)/(-4.91 m/s²) = 6.50 s²
t = 2.55 s.

(d) Knowing the time it takes to hit the ground, we can now calculate how fast the ball is travelling when it hits the ground. Using equation 2.6, we get

v = 0 m/s + (-9.81 m/s²)(2.55 s) = 25.0 m/s.

6

 

We know that the rocket iniatally moving 50 m/s and under constant acceleration upward reaches a height of 150 m. From this information we can calculate how fast the rocket is moving upward at this point and time. Doing this we find the rocket to be moving 55.7 m/s.
a) The problem now becomes how far will a rocket moving 55.7 m/s rise under the acceleration due to gravity, a = -9.81 m/s/s, before having a velocity, v_f, = 0.
Once again applying equation 5(2ay = v_f² - v_i²) we get y = 158 m. Since the rocket was already 150 meters off the ground, the maximum height is 150 m + 158 m = 308 m.
b) It takes 2.85 s for the rocket to reach 55.7 m/s under a = 2 m/s/s. From there we need to calculate how much time is required to travel the next 158 m. Equation 3 yields t' = (0 - 55.7 m/s)/-9.81 m/s/s = 5.68 s. Thus, the total time after launch to maximum height is 2.85 + 5.68 s = 8.53 s.
c) Now the rocket will fall, from rest, through a vertical displacement of 308 meters.
v_i = 0, y = -308 m, a_y = -9.81 m/s/s
Equation 4 (y = v_it + (1/2)a_yt²) can be solved to yield
t = 7.92 seceonds to fall
therefore, total time in air equals 8.53 s + 7.92 s = 16.5 seconds.