PEP 294: Solutions

X. Linear Kinematics

Introductory Problems (pp. 328-329)

1. Track length = 400 m, Dt = 12 min = (12)(60) s = 720 s, distance = 6.5 laps = (6.5)(400) m = 2600 m

a. Actual distance:

d = 2600 m

b. Displacement:

d = (-160, 0) m

c. Average speed:

v = = = 2600 / 720 = 3.61 m/s

d. Average velocity:

v = = = = (-0.22, 0) m/s

2. a = -0.5 m/s², Dt = 7s. Compute v₁.

Tip -- Although not stated explicitly, the final velocity is 0 (v₂ = 0 m/s).

v₁ = 0 - (-0.5)(7) = 3.5 m/s

 

3. v₂ = 5 m/s, Dt = 1.5 s, a = 3 m/s². Compute v₁.

v₁ = 5 - (3)(1.5) = 0.5 m/s

4. d₁ = 400 m (E), d₂ = 500 m (45° NE). Compute d.

- From the law of cosines:

· From the law of sines:

5. v₁ = 5 m/s (N), Dt₁ = 120 s, v₂ = 4 m/s (W), Dt₂ = 180 s. Compute d.

- Compute d₁ first:

d₁ = (5)(120) = 600 m (N)

- Compute d₂ using the same method:

d₂ = 500 m (W)

- Compute resultant d (treat d₁ and d₂ as the components of d since d₁ and d₂ are perpendicular to each other):

q = tan^-1(1.20) = 50.19° (NW)

6. The gravitational acceleration affects the vertical motion of a projectile, but not the horizontal motion. So the vertical and horizontal motion components show different trends of motion. This is why we analyze the vertical and horizontal motions separately.

7. v_x = 5 m/s, v_y = 3 m/s

a. There is no acceleration in horizontal motion, so the horizontal velocity does not change. Therefore, the horizontal velocity 0.5 s into its flight = 5 m/s.

b. Again, the horizontal velocity does not change through the entire period of flight. The horizontal velocity midway through its flight is the same to the initial velocity (5 m/s).

c. The ball’s horizontal velocity immediately before contact with the ground is again 5 m/s.

d. The ball’s vertical velocity at the apex is always 0 m/s.

e. The ball’s vertical velocity midway through its flight is 0 m/s since the ball’s trajectory shows a symmetric path so that it reaches the apex midway through its flight.

f. The ball’s vertical velocity immediately before contact with the ground is the same to the initial vertical velocity in magnitude, but opposite in direction. So the vertical velocity is -3 m/s.

8. They both hit the ground at the same time since both are under the influence of gravitational acceleration. Gravitational acceleration is always 9.81 m/s² (downward) regardless of the shape and mass of the object.

9. v = v_x = 22 m/s, Dt = 0.7 s.

- Formulas for the horizontal motion:

(1) v_h = v_x

(2) d_h = v_x^.t

- From Formula (2) above:

d = (22)(0.7) = 15.4 m

10. v = v_y = 9.2 m/s; Compute H.

- Formula for the height of the apex is:

- Height of the apex:

Additional Problems (pp. 329-330)

2. See IP 4 above.

3. See IP 5 above.

4. v_current = 0.5 m/s (S), v_wind = 0.7 m/s (W), Dt = 5 min = 300 s

- Compute the resultant velocity first:

54.46 ° (SW)

- Compute the displacement:

d = v^.Dt

d = v^.Dt = (0.86)(300) = 258.0 m (54.46° SW)

5. v_wind = 4 m/s (W), v_current = 2 m/s (45° NE), Dt = 10 min = 600 s; Compute d.

- First compute the resultant velocity using the law of cosines:

- Compute the direction angle using the law of sines:

= 0.479

q = sin^-1(0.479) = 28.62° N of due W or 61.38° NW

- Now, compute the displacement:

d = v^.Dt

d = v^.Dt = (2.95)(600) = 1770.0 m (61.38° NW)

6. 1 yd = 0.91 m, d_Cowboy = 50 yd = (50)(0.91) = 45.5 m, v_Cowboy = 8 m/s

- First figure out the distance the Buffalo Bill should run to catch the Dallas Cowboy:

- So the ratio of d_Bill to d_Cowboy is:

In other words, the Bill should run 1.04 times faster than the Cowboy.

- Apply the ratio to velocity:

v_Bill = (v_Cowboy)(1.04) = (8)(1.04) = 8.32 m/s

So, the Bill should run at a speed of 8.32 m/s to catch the Cowboy who’s carrying the ball.

7. q = 45° , 2T (flight time) = 3 s, T (ascending time) = 1.5 s; Compute H.

Tip -- Both the height of the apex and the flight time are the functions of the initial vertical velocity. So we need to compute the initial vertical velocity first. Compute the initial vertical velocity from the ascending time (T), and then the height of apex from the initial vertical velocity:

- From the formula for ascending time (T):

v_y = (1.5)(9.81) = 14.72 m/s

- From the formula for H:

8. D = 45.8 m, H = 24.2 m, 2T = 4.4 s; Determine if q > 45°.

Tip -- Compute the initial vertical velocity (v_y) and the initial horizontal velocity (v_x), and see if v_y is larger than v_x.

- Compute v_y:

v_y = (9.81)(2.2) = 21.58 m/s

- Compute v_x using the formula for D:

D = (v_x)(2T);  45.8 = (v_x)(4.4)

- v_y > v_x: therefore q > 45°

9. q = 35° , v = 10 m/s; Compute H & D.

- First resolve the projection velocity into components:

v = ( v_x, v_y ) = ( v^.cos q, v^.sin q ) = ( (10)(cos 35°), (10)(sin 35°) ) = ( 8.19, 5.74 ) m/s

- Compute H:

· To compute D, first you need to compute the ascending time (T) and eventually the flight time (2T):

Filght time = 2T = (2)(0.59) = 1.18 s

· Compute D:

D = v_x^.(2T) = (8.19)(1.18) = 9.66 m

10. v = 45 m/s, q = 10°; Compute D.

- Resolve the projection velocity vector into components:

v = ( v_x, v_y ) = ( v^.cos q, v^.sin q ) = ( (45)(cos 10°), (45)(sin 10°) ) = ( 44.32, 7.81 ) m/s

- Compute the ascending time and eventually the flight time:

Flight time = 2T = (2)(0.80) = 1.60 s

- Compute D:

D = v_x^.(2T) = (44.32)(1.60) = 70.91 m