Vector Kinematics

Two important things to keep in mind - keep components separate from each other in a kinematics problem. Second, don't subsitute values until you are ready to get an answer. Keeping things algebraic makes things much cleaner.

Handout Solutions:
5.
a) delta_t = v₀ sin(theta) / g
b) delta_y = (v₀ sin(theta))^2/2g

6. displacement is 65.9 km at [E 15.6 degrees N][E 17.0 degrees N]

7. x-component: 27.19 m/s, y-component: 12.67 m/s

8. 43.01 m/s at 54.46 degrees above the x-axis

9.

10. 60 mi/h 34.28 mi/h

11.
a) 1.72 seconds
b) 14.58 meters
c) double the answer in (a) of 3.44 seconds
d) 124.7 meters

12. Check your notes for these definitions.

Book problems:

Question 10. No - the magnitudes may be the same, but the directions are different. This means the velocities are not the same.

9.
a) West component: 785 km/h* sin(38.5 degrees) = 488.7 km/h (Note that this would be negative according to our usual coordinate system of positive to the right.)
North component: 785 km/h* cos(38.5 degrees) = 614.3 km/h
b)
West: 1466.1 km
North: 1842.9 km

16.
a) Vertical component of acceleration = 3.8 m/s^2 * sin(30) = 1.9 m/s^2
b) delta_y = v0y*delta_t + 1/2*a_y*delta_t^2, but since she starts from rest, v0_y = 0. The delta_y is the elevation change of 335 meters.

Thus delta_t = (2*delta_y/a_y)^1/2 = 18.77 seconds