Solutions to Week 4 Assignment

 

  1. (a) To return home, the pilot must travel opposite his previous day's displacement. Thus, he musts travel 55 mi east and 25 mi north.
         Find the distance.
          distance = Ö(55mi)2 + 25mi)2 = 60mi
         Find the direction.
         q = tan-1(25mi/55mi) = 24° north of east.
         So, the pilot must travel 60mi at 24° north of east.
    (b) The pilot has flow 55mi + 25mi + 60mi = 140mi extra miles.
  2. (a) bx = 7.1cos(-14°) = 6.9.
         by = 7.1sin(-14°) = -1.7.
    (b) |c| = Öcx2 + cy2 = Ö(-1.8)2 + (-6.7)2 = 6.9.
         q = tan-1(-6.7/-1.8) = 15° CW from the -y-axis.
    (c) |c + b| = Ö(cx + bx)2 + (cy + by)2 = Ö(-1.8 + 6.9)2 + (-6.7 -1.7)2 = 9.8
         q = tan-1(-8.4/5.1) = 31° CCW from the -y-axis.
    (d) x-component = cx - bx = -1.8 -6.9 = -8.7
         y-component = cy - by = -6.7 + 1.7 = -5.0
  3. Let the +y-direction be up, and the +x-direction be to the right.
    (a) T1 = the tension in the rope from which the pulley hangs
         T2 = the tension in the rope fastened to the wall
         Use Newton's second law.
         SFy = T1sinq  - Mg = 0
         SFy = T1cosq - T2 = 0
         The tension in T2 is due to the mass M, so T2 = Mg.
         So, T1cosq  = Mg, and  T1sinq  = Mg.
         According to these equations, cosq  = sinq , which is true q  = 45° for 0° < q  < 90°.
         Thus, T1 = M/cos45° = Ö2Mg.
    (b) As found in (a) q = 45°.
  4. (a) v = C/4/Dt = C/4Dt = 2pr/4Dt = pr/2Dt = p(10.0m)/2(1.60s) = 9.82m/s.
    (b) Let east be the +x-direction and north the +y-direction.
         Dv = Öv2 + v2 = Ö2v2 = vÖ2
        q = tan-1-v/v = tan-1(-1) = 45° south of east (SE)
         Dv = Ö2(9.82m/s) southeast = 13.9m/s southeast.
    (c) aav = Dv/Dt = 13.88m/s southeast/1.60s = 8.68m/s2 southeast.
  5. (a) At the maximum height, vy = 0.
         vy = v0sinq  - gt = 0
         So, t = v0sinq/g.
         Find the maximum height
         Dy = (v0sinq)t - 1/2gt2
              = v0sinq(v0sinq/g) - 1/2g(v0sinq/g)2
              = v02sin2q/g - v02sin2q/2g
              = v02sin2q/2g
              = (22.0m/s)2sin260.0°/2(9.8m/s2)
              = 19m higher than where it was hit.
    (b) The elapsed time is twice that found in part (a)
          t = 2v0sinq/g = 2(22.0m/s)sin60.0°/9.8m/s2 = 3.9s.
    (c) x = v0xt = v0cosq t = (22.0m/scos60.0°(3.9s) = 43m.
  6.  vf2 - v02 = 2aDxf
      vf2 - 0 = 2aDxf
      vf2 = 2aDxf
     and 
     (1/2vf)2 = 2aDx = 1/4vf2
    Form a proportion.
    vf2/1/4vf2 = 2aDxf /2aDx 
      4 = Dxf /Dx
      Dx= Dxf/4 = 2.0m/4 = 0.50m.
  7. (a) t = xacross /vboy and vwater = xdownstream/t, so 
        vwater = xdownstream/(xacross /vboy) = (xdownstream/ xacross)(vboy) = (50.0m/25.0m)(0.500m/s) = 1.00m/s.
    (b) Use the Pythagoream theorem.
          vbf = Ö(0.500m/s)2 + (1.00m/s)2 = 1.12m/s.
  8. Let +x be east and +y be north.  
      x = x0 + v0xt + 1/2axt2 = 2.0m + (0)(2.0s) + 1/2(5.0m/s2)(2.0s)2 = 12m
      y = y0 + v0yt + 1/2ayt2 = 0 + (20m/s)(2.0s) + 1/2(0)(2.0s)2 = 40m
    So, r = 12m east and 40m north.