56. The route followed by a hiker consists of three displacement vectors 25.0° n

Question

56. The route followed by a hiker consists of three displacement vectors 25.0° north of east. Vector B is not along a measured trail, but the hiker uses a compass and knows that the direction is 41.0 east of south. Similarly, the direction of vector C is 35.0 north of west. The hiker ends up back where she started. Therefore, it follows that the resultant displace- ment is zero, or A + B + C = 0, Find the magnitudes of (a) vector B and (b) vector C 14. As a tennis bal is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?

Explanation / Answer

East = +i
North = +j

B = 41 east of south = 270+41 = 311 degrees
C = 35 north of west = 180-35 = 145 degrees

A = 1550*cos(25) i + 1550*sin(25)j
B = B*cos(311) i + B*sin(311)j
C = C*cos(145) i + C*cos(145)j

A+B+C = 0 ==> ai + bi + ci = 0 and aj + bj + cj = 0

1550*cos(25) + B*cos(311) + C*cos(145) = 0
1540*sin(25) + B*sin(311) + C*sin(145) = 0

We have 2 equations with 2 unknowns

1404.77+0.6560B-0.8191C = 0...........equation 1

650.83-0.7547B-0.5736C = 0.............equation 2

Now multiply equation 2 by 0.6560 and equation 1 by 0.7547 and then add them

1060.18+0.4951B+0.6182C = 0........equation 3

426.94-0.4951B-0.3763C = 0..........equation 4

So by adding equation 3 and 4 we get,

1487.12+0.2419C = 0

So, C = 6147.66 m

Put this value in equation 2, we get

B = 3810 m

So, magnitude of vector B and vector C is,

B = 3810 m

C = 6147.66 m

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