1.A pinball bounced around its machine before resting between two bumpers. Befor

Question

1.A pinball bounced around its machine before resting between two bumpers. Before the ball came to rest, its displacement (all angles measured counterclockwise from the horizontal x axis) was recorded by a series of vectors. 83 cm at 90° 55 cm at 145° 69 cm at 227° 43 cm at 281° 61 cm at 27° Use the figure on the right to draw these displacements in order with the first vector starting at the large "X", and then determine the total displacement's magnitude and direction. 2.An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s

2. If the wind's acceleration lasts for 2.00 s, find the magnitude r and direction (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.)

3.Now, assume the same bird is moving along again at 4.00 mph in an easterly direction but this time the acceleration given by the wind is at a 48.0 degree angle to the original direction of motion. If the magnitude of the acceleration is 0.400 m/s2, find the displacement vector , and the angle of the displacement, 1. Enter the components of the vector and angle below. (Assume the time interval is still 2.00 s.)

Explanation / Answer

1)

You're supposed to measure the distance from X to the end of vector 5 using the appropriate scale, and measure the angle (counterclockwise from X) using a protractor.

Mathematically:

x = [83cos90+55cos145+69cos227+43cos281+61cos27] cm = -29.56 cm

y = [83sin90+55sin145+69sin227+43sin281+61sin27] cm = 49.56 cm

so d = (x² + y²) = 57.71 cm

and = arctan(49.56/-29.56) = -59.19º +180º (to get into QII) = 120.8º

Of course, anything close by ruler and protractor would be fine.

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