ANSWER TO CHAPTER 3 HOMEWORK

1.  (a)    V1x = -6.0,   V1y = 0;

              V2x = V2cos45° = 4.5cos45° = 3.18 =  3.2,

              V2y = V2sin45° = 4.5sin45° = 3.18 = 3.2.

         (b) For the components of the sum we have

               Rx = V1x + V2x = -6.0 + 3.18 = -2.82;

               Ry = V1y + V2y = 0 + 3.18 = 3.18.

              We find the resultant from

               R = (Rx2 + Ry2)1/2 = [(-2.82)2 + (3.18)2]1/2 = 4.3;

               tan q = Ry/Rx =(3.18)/(2.82) = 1.13, which gives

               q = 48° above -x-axis.

            Note that we have used the magnitude of Rx for the angle indicated on the diagram.

 
2. (a) For the components we have

               Rx = Ax - Bx + Cx 

                    = 44.0cos28.0° - (-26.5cos56.0°) + 0 = 53.7;

               Ry  = Ay - By + Cy 

                    = 44.0sin28.0° - 26.5sin56.0° - 31.0 = -32.3.

          We find the resultant from

               R = (Rx2 + Ry2)1/2 = [(53.7)2 + (-32.3)2]1/2 = 62.7;

               tan q = Ry/Rx = (32.3)/(53.7) = 0.602, which gives

               q = 31.0° below +x axis.

           Note that we have used the magnitude of Ry for the angle indicated on the diagram.

          (b) For the components we have

                Rx = Ax + Bx -Cx 

                    = 44.0cos28.0° + (-26.5cos56.0°) - 0 = 24.0;

                Ry = Ay + By - Cy

                     = 44.0sin28.0° + 26.5sin56.0° - (-31.0) = 73.6.

            We find the resultant from

                R = (Rx2 + Ry2)1/2 = [(24.0)2 + (73.6)2]1/2 = 77.4;

               tan q = Ry/Rx = (73.6)/(24.0) = 3.07, which gives

               q = 71.9° above +x axis.

         (c) For the components we have

                Rx = Cx - Ax - Bx

                     = 0 - 44.0cos28.0° - (-26.5cos56.0°) = -24.0;

                Ry = Cy - Ay - By

                     = -31.0 - 44.0sin28.0° -26.5sin56.0° = -73.6.

               We find the resultant from

                  R = (Rx2 + Ry2)1/2 = [(-24.0)2 + (-73.6)2]1/2

                            = 77.4;

                         tan q = Ry/Rx = (73.6)/(24.0) = 3.07,

                which gives

                   q = 71.9° below -x axis.

                  

 

3. (a) Because we do not know the displacement over the given time interval, the average velocity is unknown.

             (b) The average acceleration is

                       aav = Dv/Dt =[(27.5m/s)i -(-18.0m/s)j]/(8.00s0 = (3.44m/s2)i + (2.25m/s2)j

                  the magnitude is is

                       | aav | = [(3.44m/s2)2 + (2.25m/s2)2]1/2 = 4.11m/s2

                       We find the direction from

                       tan q = (2.25m/s2)/(3.44m/s2) = 0.654, which gives  q = 33.2° north of east. 

             (c)  Because we do not know the distance traveled, the average speed is

                       unknown. 

4. We choose a coordinate system with the origin at the release point, with x horizontal      and y vertical, with the positive direction down. We find the time of fall from the vertical displacement:

                   y = yo + voyt + 1/2ayt2

                   9.0m = ) + 0 + 1/2(9.80m/s2), which gives t = 1.35s.

       The horizontal motion will have constant velocity.

       we find the initial speed from ;

                    x = xo + voxt

                   8.5m = ) + vo(1.35s), which gives  vo = 6.3m/s.

  

5.    (a) At the highest point, the vertical velocity vy = 0. We find the maximum height h from

                     vy2 = voy2  + 2ay(y - yo);

                     0 = [(51.2m/s)sin44.5°]2 + 2(-9.80m/s2)(h - 0), which gives h = 65.7m.

        (b) Because the projectile returns to the same elevation, we have

                     y = yo + voyt + 1/2ayt2;

                     0 = 0 + (51.2m/s)(sin44.5°)t + 1/2(-9.80m/s2)t2, which gives t= 0, and 7.32s.

               Because t = 0 was the launch time, the total time in the air was  7.32s.

         (c) We find the horizontal distance from

                     x = voxt (51.2m/s)(cos44.5°)(7.32s) =  267m.

         (d) the horizontal velocity will be constant; vx = vox (51.2m/s)cos44.5° = 36.5m/s.

                we find the vertical velocity is

                    vy = voy + ayt = (51.2m/s)sin44.5° + (-9.80m/s2)(1.50s) = 21.2m/s.

                The magnitude of the velocity is

                     v = (vx2 + vy2)1/2 = [(36.5m/s)2 + (21.2m/s)2]1/2 = 42.2m/s

                We find the angle from

                     tan q = vy/vx = (21.2m/s)/(36.5m/s) = 0.581, which gives   q = 30.1° above the horizontal.

6.   The speed of the speck is

                    v = 2pr/T  = 2prf.

          Thus the centripetal acceleration is

                    aR = ( v2/r ) = (2prf)2/r  = 4p2rf2    

                         = 4p2(0.15m)[(45min-1)/960s/min)]2 = 3.3m/s2

7.  If       vPA is the velocity of the airplane with respect to the air,

                   vPG is the velocity of the airplane with respect to the ground, and

                   vAG is the velocity of the air (wind) with respect to the ground, then

                   vPG = vPA + vAG, as shown in the diagram.

          (a)  From the diagram we find the two components of vPG:

                              vPGE = vAGcso45° = (90.0 km/h)cos45° = 63.6 km/h;

                      vPGS = vPA - vAGsin45° = 550 km/h - (90.0 km/h)sin45° = 486km/h.

                 For the magnitude we have

                      vPG = (vPGE2 + vPGS2)1/2 = [(63.6 km/h)2 + (486 km/h)2]1/2 =  490 km/h.

                 We find the angle from 

                      tan q = vPGE/vPGS = (63.6 km/h)/(486 km/h) = 0,131, which gives

                         q = 7.45° east of south.

            (b) Because the pilot is expecting to move south, we find the easterly distance 

                  from this line from

                     d = vPGEt (63.6 km/h)(12.0 min)(760 min/h) =  12.7km.

                 Of course the airplane will also not be as far south as it would be without the wind.

                    

                 

  
*8 The position is r = (6.0m)cos(3.0s-1)ti + (6.0m)(3.0s-1)tj.

(a) We find the velocity by differentiating:

                  v = dr/dt = -(6.0m)(3.0s-1)sin(3.0s-1)ti + (6.0m)(3.0s-1)cos(3.0s-1)tj

                    = - (18.0m/s)sin(3.0s-1)ti + (18.0m/s)cos(3.0s-1)tj.

(b) We find the acceleration by differentiating:

                a = dv/dt = - (18.0m/s)(3.0s-1)cos(3.0s-1)ti - (18.0m/s)(3.0s-1)sin(3..0s-1)tj

                   = - (54.0m/s2)cos(3.0s-1)ti - (54.0m/s2)sin(3.0s-1)tj.

(c) The magnitude of r is 

               |r| = {[(6.0m)cos(3.0s-1)t]2 + [(6.0m)sin(3.0s-1)t]2}1/2 = 6.0m.

Thus thus particle is always 6.0m from the origin, so it is traveling in a circle.

(d) We see that

               a = - (9.0s-2)r, so we have

               a = (9.0s-2)r,

      with the angle between the vectors being 180°, that is in opposite directions.

(e) We see that

              |v| = {[(18.0m/s)sin(3.0s-1)t]2 + [(18.0m/s)cos(3.0s-1)t]2}1/2 =  18.0m/s, so 

              v = (3.0s-1)r, and 

              a = (9.0s-2)r = v2/r.

*9. We choose a coordinate system with the origin at the base of the hill, with x horizontal and y vertical, with the positive direction up.

 When the object lands on the hill, y = xtanf .

The distance up the hill is given by

              d2 = x2 + y2 = x(1 + tan2f ).

Thus to maximize d, we can maximize x.

For the horizontal and vertical motions we have

             x = x0 + v0xt = 0 + (v0cosq )t =  (v0cosq )t;

             y = y0 + v0yt + 1/2ayt2 = 0 + (v0sinq)t + 1/2(-g)t2  = (v0sinq)t - 1/2gt2 .

We combine these equations to get x a function of q :

             y = xtanf  = (v0sinq)(x/v0cosq) -1/2g(x/v0cosq)2, which gives

             x = (2v02cosq/g)(sinq - cosq tanf).

We can simplify this by using trigonometric identities to get functions of 2q.

             x = (v02/g)[sin2q - (1 + cos2q)tanf].

To find the value of q  for maximum x, we set dx/dq  = 0:

            dx/dq  = (v02/g)[2cos2q - (-2sin2q)tanf] = 0, which gives

             tan 2q  = -cotf, or q = 1/2tan-1(-cotf).

Note that the negative sign means using the angle greater than 90° for the inverse tangent.

 

 

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