(1) 3.3x103 mi/hr
(2) 5.4x1011 m
(3) 2.5 ft/s
(4) a: about 3 ft/s; b: velocity is about -4 ft/s so speed is about 4 ft/s
(5) -2 m/s2
(6) 80 ft/s
(7) a: about 3 m/s2; b: a=0. (slope is zero)
(8) Yes, the velocity becomes zero at about 68 s. The acceleration is never zero; it is constant and negative (about -2/3 m/s2).
Homework #5:
(1) a: a = 2.5 m/s2; b:
x = 278 m or about 280 m (sig figs)
(2) a: a = -1.67 m/s2; b:
x = 7.5 m
(3) a: a = -8.8 ft/s2; b:
t = 10 s
(4) v = 18.9 m/s2
(5) y = 20.4 m
(6) a: the time from top to bottom is 1.93 s, so the total time is twice that, or 3.9 s
b: The ball was thrown upward at 18.9 m/s.
c: Just before hitting the ground, the ball's velocity is 18.9 m/s downward, or -18.9 m/s
Homework #6
(1) a = -12 ft/s2; to slow to half the velocity, vf is now 27 ft/s, using vf = vi + a(
t), getting t = 2.2 s
(2) vi = 20 ft/s = 6.1 m/s. Use yf = yi + vi(
t) - (1/2)g(t)2, with t = 1.2s to get yf = 0.26 m. Find the velocity using vf = vi - g(t) and get vf = -5.7 m/s. (It is negative, so the ball is on the way down, almost to its starting height).
(3) To add the velocities, draw the helicopter's velocity straight up, then draw the wind velocity head-to-tail from the first vector, forming a right angle in this case. The vector sum is the hypotenuse of the triangle, with a magnitude of 3.6 m/s (from the Pythag. thm) and a direction of 56° from horizontal.
(4) Again, to add vectors, draw them head-to-tail and the resultant vector goes from the tail of the first vector to the head of the last vector. Here, the boat's velocity relative to the river does not point along an axis, so you have to find the x- and y-components first. The river flows directly eastward, so its x-component is 3 mi/hr and its y-component is 0. The boat's x-component is vb.rel-to.r.,x = -7.07 mi/hr, and vb.rel-to.r.,y = 7.07 mi/hr. We get the x-component of the resultant by adding the x-components of the vectors: vb.rel-to.g,x = (-7.07 + 3) mi/hr = -4.07 mi/hr. The y-component of the resultant is 7.07 mi/hr. Combine the components to get magnitude and direction of 8.2 mi/hr at 30° west of north.
(5) Use the range equation to fine 144 m.
(6) It goes 32 m high, and the range is 127 m.
(7) Solve the range equation for
= (1/2) arcsin(g R/vi2). The first angle you get is 32.1° The second angle is the complementary angle (see the book!) 90° - 32.1° = 57.9° (Note that we did not cover this in class, so it will not be covered by the exam, but that doesn't mean it's not worth knowing!)