Physics 101 - Homework # 2 Solutions

• Problem 1-50
• Problem 3-2
• Problem 3-13
• Problem 3-20
• Problem 3-30
• Problem 3-36
• Problem 3-43
• Problem 3-48
• Problem 3-55
• Problem 3-60

Chapter 1

Problem 50)

a) A + B + C + D   =   (-2i - 3j) + (i + 2j + 3k) + (3j + 3k) + (-2i -k)

= - 3i + 2j + 5k

b) A - D   =   (-2i - 3j) - (-2i -k) = -3j + k

c) A + D - B   =  (-2i - 3j) + (-2i -k) - (i + 2j + 3k) = - 5i - 5j -4k

d) A - C   =   (-2i -3j) - (3j + 3k) = -2i - 6j -3k

|A - C| = [(-2)² + (-6)² + (-3)²]^1/2 = 7

Chapter 3

Problem 2)

Given r = (c₁ - c₂t)i + (d₁ + d₂t + d₃t²)j

= [(12 m) - 3 m/s)t]i + [(-20 m) + (-3 m/s)t + (0.5 m/s²)t²]j

To find the time to pass through   x=0   we solve the x-component equation:

x = 0 = (12 m) - (3 m/s)t    to get   t=4 s

Setting   x = y    gives us:

(12 m) - ( 3 m/s)t = (-20 m) - (3 m/s)t + (0.5 m/s²)t²

0 = 32 m - (0.5 m/s²)t²

t² = 64 s²

from which we get     t = ± 8 s

To find the locations, we substitute these values into the expression for r:

t = - 8 s:   r = (36i + 36j) m

t = + 8 s:   r = (-12i - 12j) m

Problem 13)

From the definition of instantaneous acceleration, we have

a = dv/dt = d(2.2 m/s i + 3.7 m/s²t j + [(3.3 m/s) - 1.2 m/s³)t²]k]/dt

we differentiate each component separately, and then add them:

= (3.7 m/s²)j - (2.4 m/s³)t k

Note: the textbook has a typo in the solution in the back; it has - (2.4 m/s²)t for the second term, which is not correct dimensionally.

Problem 20)

We are given v₀ = ( 50 m/s)j and a = [35 m/s² + (2 m/s⁵)t³]i + [4 m/s² - (1 m/s⁴)t²]j

To find the velocity, we need to integrate this with respect to time;

v = a dt

which gives

v - v₀ = [(35 m/s²)t + (2 m/s⁵)t⁴/4]i + [(4 m/s²)t - (1 m/s⁴)t³/3]j

or

v = [(35 m/s²)t + (0.5 m/s⁵)t⁴]i + [ 50 m/s + (4 m/s²)t - (1 m/s⁴)t³/3]j

To find the position, we must integrate again :

r = v dt

This gives

r = [(35 m/s²)t²/2 + (0.5 m/s⁵)t⁵/5]i + [(50 m/s²)t + (4 m/s²)t²/2 - (1 m/s⁴)t⁴/12]j

substituting t= 3 s into the above expressions for v and r, we have

v = (146 m/s)i + (53 m/s)j

r = (182 m)i + (161 m)j

Problem 30)

First choose a coordinate system: x is horizontal, y is vertical, and the origin is the base of the building.

a) Use the vertical motion to get the time of flight:

y = y₀ + v_0yt + (1/2)a_yt²

Here we have y=0 (when it hits the ground), an initial height of y₀ = 10 m, v_0y = 0 and a_y = = -9.8 m/s², so

0 = 10m + 0 + (1/2)(-9.8 m/s²)t²

Solving for t we get

t = 1.43 s

Now we substitute this time into the horizontal position equation

x = x₀ + v_0xt

(there is no acceleration in the x-direction), or

8.0 m = 0 + v_0x(1.43 s)

thus

v_0x = 5.6 m/s

b) If the ball is thrown up at an angle, the initial velocity will need to change; now,

y = y₀ + v_0yt + (1/2)a_yt²

becomes

0 = 10m + v₀(sin29°)t + (1/2)(-9.8 m/s²)t²

and the x equation becomes

x = v_0xt

8.0 m = v₀(cos29°)t

These are two equations in two unknowns, so we can solve them. Solving the second equation for v₀ in terms of t, and sticking this in the first equation gives

0 = 10m + (8.0 m)/cot(29°) + (1/2)(-9.8 m/s²)t²

solving this for t gives

t = 1.7 s

and plugging this back into the second equation gives

v₀ = 5.4 m/s

c) from above, t = 1.7 s

Problem 36)

We will use a coordinate system with the origin at the base of the tower, x=horizontal, and y=vertical, so   y₀ = H   where H is the tower's height.

a) For the horizontal motion:

x = v_0xt = v₀cos45°t

20 m = v₀(0.707)(4 s) which gives

v₀ = 7.1 m/s

b) For the vertical motion:

y = y₀ + v_oyt + ½a_yt²

0 = H + (7.1 m/s)(cos 45°) + ½(-9.8 m/s²(4 s)²

which gives

H = 58 m

c) To find the components of the velocity:

v_x = v_0x = (7.1 m/s)cos45° = 5.0 m/s

and

v_y = v_0y + a_yt

= (7.1 m/s)(sin45° + (-9.8 m/s²)(4 s) = -34 m/s

The speed is the magnitude of the velocity:

v = (v_x² + v_y²)^1/2 [(5.0 m/s)² + (-34 m/s)²]^1/2 = 34.4 m/s

Problem 43)

The radius of the circular path is R = R_Earth + h

= (6.37 x 10³ km) + 220 km = 6.59 x 10³ km

a) The speed is   v = 2R/T

= 2(6.59 x 10⁶ m)/(89 min)(60 s/min) = 7.8 x 10³ m/s = 28 x 10³ km/h

b) The acceleration is   a = v²/ R

= (7.8 x 10³ m/s)²/(6.59 x 10⁶ m) = 9.1 m/s²    toward the earth's center.

Aside: this is suspiciously close to "g", but not exactly the same - we will see why in Chapt. 12.

Problem 48)

The velocity of a point on the rim of the flywheel is just given by

v = (4000 rev/min)(2(0.10 m)/rev)(1 min/60 s)

= 41.9 m/s

since each revolution is a path length of 2r. The centripetal acceleration is

a = v²/r

= (41.9 m/s)²/(0.10 m) = 1.75 x 10⁴ m/s² = 1790 g

since g= 9.8 m/s².

Problem 55)

This is a relative velocity problem. If we use a coordinate system with x east and y north, the velocity of the boat with respect to the water is

v_B = v_A - u,    where

v_A = 15 i km/h

   is the velocity of the boat with respect to the land and

u = 5 j km/h     is the velocity of the Gulf Stream with respect to the land.

Thus

v_B = (15i - 5j) km/h = 15.8 km/h, 18.4° south of east.

Problem 60)

a) Another relative velocity problem. In order to fly due north, the airplane must head northwest (to counterbalance the wind) as shown in the diagram. Let v_p be the velocity of the plane with respect to the air, v_g be the velocity of the plane with respect to the ground, and v_W the velocity of the wind with respect to the ground. Then

v_g = v_p + v_W

The heading angle (see diagram) is given by

sin = v_w/v_p = (65 mi/h)/(180 mi/h)

which gives

= 21.2° west of north.

Because the diagram we used to find the angle was a velocity diagram, it is independent of distances, so the angle would be the same regardless of the distance.

b) Again, from the diagram, we have

v_g = v_pcos = (180 mi/h)cos21.2° = 168 mi/h

thus the time of flight is

t = d/v_g = (673 mi)/(168 mi/h) = 4.0 h

c) If the airplane heads north until it reaches the latitude of St. Louis, its velocity with respect to the ground will be to the northeast, as shown in the diagram.

The time for the first leg of the trip (before the plane turns west) can be found from

t₁ = 673 mi/(180 mi/h) = 3.74 h

The plane must then travel towards the east a distance

d = v_Wt₁ = (65 mi/h)(3.74 h) = 243 mi

since this is the distance the wind blew the plane westwards. As it goes west, its speed with respect to the ground is given by

v_new = v_p - v_W = 180 mi/h - 65 mi/h = 115 mi/h

Thus the time required is

t₂ = 243 mi/(115 mi/h) = 2.11 h

The total time for the trip is then

t₁ + t₂ = 3.74 h + 2.11 h = 5.85 h

which is substantially longer than using the flight plan in part a).