A girl delivering newspapers covers her route by traveling 3.00 blocks west, 6.0

Question

A girl delivering newspapers covers her route by traveling 3.00 blocks west, 6.00 blocks north, and then 6.00 blocks east. (a) What is her resultant displacement? blocks at ° north of east (b) What is the total distance she travels?

2. A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are d1 = 4.10 m to the north, d2 = 2.20 m northeast, and d3 = 1.10 m at = 35.0° west of south (figure below). Starting at the same initial point, an expert golfer could make the hole in what single displacement?

3.A snow-covered ski slope makes an angle of slope = 36.0° with the horizontal. When a ski jumper plummets onto the hill, a parcel of splashed snow is thrown up to a maximum displacement of 1.70 m at snow = 17.0° from the vertical in the uphill direction as shown in the figure below.

(a) Find the component of its maximum displacement parallel to the surface.
m

(b) Find the component of its maximum displacement perpendicular to the surface.
m

Explanation / Answer

a) Take west as positive x direction and north as positive y direction.
East is opposite of east. Thus east is negative x direction.

Therefore, displacement in x direction = X= 3 blocks - 6 blocks = -3 blocks
Displacement in y direction = Y = 6 blocks
Magnitude of resultant displacement = sqrt(X^2 + Y^2)
= sqrt(-3^2 + 6^2) blocks
= sqrt(9 + 36) blocks
= sqrt(45) blocks
= 6.71 blocks
Let the resultant displacement make angle theta with x axis.
theta = atan(Y/X) = atan(6/-3) = atan(-2) = -63.43 deg =180-63.43 = 116.57 deg
Positive x direction is towards west and positive y direction is towards north. Thus angle 116.57 deg calculated is north of west.

Ans: Resultant displacement is 6.71 blocks north of west

b)Total distance = 3 blocks + 6 blocks + 6 blocks
= 15 blocks

3(a) x = 1.7m * sin(17º+36º) = 1.36 m
b) y = 1.7m * cos(53º) = 1.02 m

4)  I'm assuming that the 112N force is pulling at 50o above the x axis and the 100.3N force is at 59o below the x axis.

So using components F1x = F1*cos(50) = 112N*cos(50) = 71.99N
F1y = F1*sin(50) = 112N*sin(50) = 85.80N

For F2 we have F2x = 100.3*cos(-59) = 51.66N
and F2Y = 100.3N*sin(-59) = -85.97N

So Fx = F1x + F2x = 71.99 + 51.66 = 123.65N
and Fy = 85.80 - 85.97 = -0.17N

So F = sqrt(123.65^2 + -0.17^2) = 123.65 N
and = arctan(Fy/Fx) = arctan(-0.17/123.65) = 179.92o

b) To make the net force zero a third person would pull with a force = 123.65N and at an angle od 179.92+180 = 359.92o

a) C = A+B =(11i-j-3k)m

b)D = (-5i-20j+9k)m

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