Starting at the origin, you travel a distance 5.0 m in a direction 60.8 degrees

Question

Starting at the origin, you travel a distance 5.0 m in a direction 60.8 degrees north of east. Then, from this new postition, you travel another distance 2.6 m in a direction 41.2 degrees north of east.

a) In your final postition, how far are you from the origin? (answer in m)

b) In your final position, how many degrees north of east are you as measured from the origin? (answer in degrees)

c) Suppose that during both phases of your motion, you moved with a constant speed of 9.6 m/s. How much time does the whole trip take, from the origin to your final position? (answer in s)

d) What is the magnitude of your average velocity for the whole trip from the origin to your final position? (answer in m/s)

e) What was the magnitude of your average acceleration for the whole trip from the origin to your final position? Answer in m/s^2)

Explanation / Answer

Given

  
   position vector 1 r1 = r1x i +r2y j

           =5 cos60.8 i + 5 sin 60.8 j

   and r2 = r2x i + r2y j

       = 2.6 cos 41.2 i + 2.6 sin 41.2 j

the final position of you is r = r x i + ry j

               = (5 cos60.8+2.6 cos 41.2 )i+(5 sin60.8+2.6 sin 41.2)j

               = (4.4)i + (6.1)j

a)
magnitude is R = sqrt((4.4)^2+(6.1)^2)= 7.52 m from the origin

b)

the direction is theta = arc tan (ry/rx)

           = arc tan (6.1/4.4)
           = 54.2 degrees

c) constant speed is 9.6 m/s

   speed = s/t ==> t = s/v = 7.6/9.6 s = 0.79167 s

d) average velocity is Vavg = ds/dt = 7.52/9.6 m/s = 0.7833 m/s

e) average acceleration is Aavg = dV/dT = 0.7833 /0.79167 m/s2 = 0.989427 m/s2

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.