Answers to free response questions:
1. (a) Because there is no upward motion, I will call down "positive."
Given information:
vi = 0
a = 9.81 m/s2
t = 4.0 s
To find vf, use: vf = vi + at
Substitute with units:
vf = 0 m/s + (9.81 m/s2)(4.0 s)
vf = 39.24 m/s
(b) Given information
same as above
To find d, use: d = vit + 1/2(a)(t^2)
Substitute with units:
d = 0 + .5(9.81 m/s2)(16 s2)
d = 78.38 m
2. (a) I now call up "positive" and down "negative." In order to find how high the ball rises,
I need to draw on the "hidden information" for one of the known variables. I know that when the ball
is at its highest point, its final velocity is zero. So,
Given information:
vi = 22.5 m/s
a = -9.81 m/s2
vf = 0 m/s
To fing d, use: vf2 = vi2 + 2ad
Substitute with units:
0 = (22.5 m/s)^2 + 2(9.81 m/s2)(d)
d = 25.8 m
(b) In order to find the time in the air, we again draw on the hidden information. When the ball returns to the height at
which it was launched, its final velocity is the same as the initial velocity, but in the
opposite direction (meaning it is negative 22.5 m/s) and its displacement is zero. So,
Given information:
vi = 22.5 m/s
a = -9.81 m/s2
vf = -22.5 m/s
To fing t, use: vf = vi + at
Substitute with units:
-22.5 m/s = 22.5 m/s + (-9.81 m/s2)(t)
t = 4.59 s
3. (a) This is not a free-fall problem. The acceleration here comes from a rocket engine, not gravity.
Given information:
Vi = 65 m/s
vf = 162 m/s
t = 10.0 s
To find d, use: d = 1/2(vi + vf)t
d = 1135 m.
(b) Given information:
same as above
To find a, use: a = (vf - vi)/t
a = 9.7 m/s2
4. (a) The graph is a best-fit line with a slope of approximately 20 m/s.
(b) The velocity is constant at 20 m/s.
5. I will call up "positive" and down "negative."
(a) Given information:
vi = 5 m/s
a = -9.81 m/s2
t = 2.0 s
To find vf, use: vf = vi + at
Substitute with units:
vf = 5 + (-9.81 m/s2)(2.0 s)
vf = -14.62 m/s
(b) Given information:
same as above
To find d, use: d = 1/2(vi + vf)t
Substitute with units:
d = 1/2(5 m/s + -14.62 m/s)(2 s)
d = - 9.62 m
The helicopter has risen 10 m in the 2 s so the bag is now 19.62 m below the helicopter.
6. Use d = average speed x time. Average speed = 10 m/s; t = 1 x 10-3. So, d = 1 x 10-2m, or
1 cm.
7. Use t = d / avg. velocity. Avg vel. = 3.5 micrometers/sec, which is 3.5 x 10-6m/s; d = 8.4 cm, which is 0.084 m.
So, t = 24,000 s, or 6.7 hrs.
8. List the given information:
vi = 20 m/s;
vf = 30 m/s;
d = 200 m.
To find a, use vf2 = vi2 + 2ad.
Substitute with units. a = 1.25 m/s2.
To find t, use vf = vi + at. Substitute with units.
t = 8 s.
9. No. The faster object may be moving at constant speed (zero acceleration).
10. List the given information:
vi = 25 m/s;
vf = 0 m/s;
t = 5 s.
To find d, use d = 1/2(vi + vf)t. Substitute with units.
d = 62.5 m.
11. List the given information:
vi = 90 km/h, which is 25 m/s;
vf = 0 m/s;
d = 0.80 m.
To find a, use vf2 = vi2 + 2ad. Substitute with units.
a = -390.6 m/s2. Note that this is a very high acceleration. It is so high, the human
body could not withstand it and would be destroyed.
12. List the given information:
a = -7.00 m/s2;
vf = 0 m/s;
d = 80 m.
To find vi, use vf2 = vi2 + 2ad.
vi = 33.5 m/s.