Practice
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**Adding vectors algebraically****- by Steven Gaskill,
2001**

**1. A football player runs directly
down the field for 35 m before turning to the right at an angle
of 25 deg from his original direction and running an additional
15 m before getting tackled. What is the magnitude and direction
of the runner's total displacement? **

Here's what you know, Dy

_{1}= -35 m, Dd = 15 m, and q = 25 deg (-115 deg from the traditional horizontal [draw it out]).Use the formula cos q= Dx / Dd and plug in: cos (-115 deg) = Dx / 15 m, so Dx = (15 m)(cos (-115 deg)) = -6.33927392611 m. Use the formula sin q = Dy

_{2}/ Dd and plug in: sin (-115 deg) = Dy_{2}/ 15 m, so Dy_{2}= (15 m)(sin (-115 deg)) = -13.5946168056 m.Add vector components ( y

_{1 }and y_{2 }in our case), so Dy = Dy_{1}+ Dy_{2}= -48.5946168056 m. Use the formula Dd = (Dx^2+Dy^2)^0.5 and plug in Dd = ((-48.5946168056 m )^2 + (-6.33927392611 m )^2)^0.5 = 49.0063585302 m = 49 m.Use the formula q = arctan ( Dy / Dx ) and plug in: q = arctan (-48.5946168056 m / -6.33927392611 m) = 82.5676116452 deg. Add 180 deg to get the proper orientation from the horizontal since this is the common rule for the arctangent of two negative vector components, so 82.5676116452 deg + 180 deg = 262.5676116452 deg. Subtract this value ( the orientation from the positive x-axis) from 270 deg to get the orientation from the -y-axis, so 270 deg - 262.5676116452 deg = 7.43238835485 deg = 7.4 deg. Leaving 49 m at 7.4 deg to the right of down field.

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**2. A plane travels 2.5 km at an
angle of 35 deg to the ground, then changes direction and travels
5.2 km at an angle of 22 deg to the ground. What is the magnitude
and direction of the plane's total displacement? **

Here's what you know, Dd

_{1}= 2.5 km, Dd_{2}= 5.2 km, q_{1}= 35 deg, and q_{2}= 22 deg.Use the formula cos q = Dx / Dd and plug in: cos (35 deg) = Dx

_{1}/ 2.5 km, so Dx_{1}= (2.5 km)(cos (35 deg)) = 2.04788011072 km, and plug in: cos (22 deg) = Dx_{2}/ 5.2 km, so Dx_{2}= (5.2 km)(cos (22 deg)) = 4.82135604375 km.Use the formula sin q = Dy / Dd and plug in: sin (35 deg) = Dy

_{1}/ 2.5 km, so Dy_{1}= (2.5 km)(sin (35 deg)) = 1.43394109088 km, and plug in: sin (22 deg) = Dy_{2}/ 5.2 km, so Dy_{2}= (5.2 km)(sin (22 deg)) = 1.94795428576 km.Add Vector components, so Dx = Dx

_{1}+ Dx_{2}= 2.04788011072 km + 4.82135604375 km = 6.86923615447 km and Dy = Dy_{1}+ Dy_{2}= 1.43394109088 km + 1.94795428576 km = 3.38189537664 km.Use the formula Dd = (Dx^2+Dy^2)^0.5 and plug in Dd = ((6.86923615447 km )^2 + (3.38189537664 km)^2)^0.5 = 7.65660640783 km = 7.7 km. Use the formula q = arctan ( Dy / Dx ) and plug in: q = arctan (3.38189537664 km / 6.86923615447 km) = 26.212167397 deg = 26 deg. Yielding 7.7 km at 26 deg to the horizontal.

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**3. During a rodeo, a clown runs 8.0
m north, turns 35 deg east of north, and runs 3.5 m. Then, after
waiting for the bull to come near, the clown turns due east and
runs 5.0 m to exit the arena. What is the clown's total
displacement? **

Here's what you know, Dd = 3.5 m, Dy = 8.0 m, Dx = 5.0 m, and q = 35 deg (55 from the normal x-axis, or north of east [ 90 deg - 35 deg = 55 deg ].

Use the formula cos q = Dx / Dd and plug in: cos (55 deg) = Dx / 3.5 m, so Dx = (3.5 m)(cos (55 deg)) = 2.00751752723 m.

Use the formula sin q = Dy / Dd and plug in: sin (55 deg) = Dy / 3.5 m, so Dy = (3.5 m)(sin (55 deg)) = 2.86703215501 m.

Add Vector components, so Dx = Dx

_{1}+ Dx_{2}= 5.0 m + 2.00751752723 m = 7.00751752723 m and Dy = Dy_{1}+ Dy_{2}= 8.0 m + 2.86703215501 m = 10.86703215501 m.Use the formula Dd = (Dx^2+Dy^2)^0.5 and plug in Dd = ((7.00751752723 m )^2 + (10.86703215501 m)^2)^0.5 = 12.9304945672 m = 13 m. Use the formula q = arctan ( Dy / Dx ) and plug in: q= arctan (10.86703215501 m / 7.00751752723 m) = 57.1843819936 deg = 57 deg Again take 90 deg and subtract 57 deg to get degrees east of north and not degrees north of east, so 90 deg - 57 deg= 33 deg. Yielding 13 m at 33 deg east of north.

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**4. An airplane flying parallel to
the ground undergoes two consecutive displacements. The first is
75 km due 30.0 deg west of north, and the secound is 155 km due
60.0 deg east of north. What is the total displacement of the
airplane?**

Here's what you know, Dd

_{1}= 75 km, Dd_{2}= 155 km, q_{1}= 30.0 deg, and q_{2}= -60.0 deg (negative since it is east of north not west of north).Use the formula cos q = Dx / Dd and plug in: cos (30.0 deg) = Dx

_{1}/ 75 km, so Dx_{1}= (75 km)(cos (30.0 deg)) = 64.9519052838 km, and plug in: cos (-60.0 deg) = Dx_{2}/ 155 km, so Dx_{2}= (155 km)(cos (-60.0 deg)) = 77.5 km.Use the formula sin q = Dy / Dd and plug in: sin (30.0 deg) = Dy

_{1}/ 75 km, so Dy_{1}= (75 km)(sin (30.0 deg)) = 37.5 km, and plug in: sin (-60.0 deg) = Dy_{2}/ 155 km, so Dy_{2}= (155 km)(sin (-60.0 deg)) = -134.233937587 km.Add Vector components, so Dx = Dx

_{1}+ Dx_{2}= 64.9519052838 km + 77.5 km = 142.451905284 km and Dy = Dy_{1}+ Dy_{2}= 37.5 km + -134.233937587 km = -96.733937587 km.Use the formula Dd = (Dx^2+Dy^2)^0.5 and plug in Dd = ((-96.733937587 km)^2 + (142.451905284 km)^2)^0.5 = 172.191753578 km = 170 km. Use the formula q = arctan ( Dy / Dx ) and plug in: q= arctan (142.451905284 km / -96.733937587 km) = -55.8209919741 deg = -55.8 deg. Multiply -55.8 deg by -1 since this is degrees south of east to get degrees north of east. Then, take 90 deg and subtract this value to get degrees east of north and not north of east, so 90 deg -(-1*(-55.8 deg)) = 34.2 deg. Yielding 170 km at 34.2 deg east of north.

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