Unit 2 Practice Problems
Answer Key
1Using a vector diagram, drawn to scale, like that shown at the right, the final displacement of the plane can be found to be . The requested displacement of the base from point B is , which has the same magnitude but the opposite direction. Thus, the answer is
.
2The displacement vectors and can be drawn to scale as at the right. The vector C represents the displacement that the man in the maze must undergo to return to his starting point. The scale used to draw the sketch can be used to find C to be .
3The total displacement is . The north and east components of this displacement are:
,
and.
4,
5After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the island. In the next 1.50 h, it travels 37.5 km due north. The components of these two displacements are:
Displacement / x-component (eastward) / y-component (northward)123 km / -61.5 km / +107 km
37.5 km / 0 / +37.5 km
Resultant / -61.5 km / 144 km
Therefore, the eye of the hurricane is now
.
6(a)
and
The resultant is .
(b)To have zero net force on the mule, the resultant above must be cancelled by a force equal in magnitude and oppositely directed. Thus, the required force is .
7The constant horizontal speed of the falcon is
.
The time required to travel 100 m horizontally is . Thevertical displacement during this time is
,
or the falcon has a vertical fall of .
8The time of flight for Tom is found from with :
.
The horizontal displacement during this time is
.
Thus, he lands .
The horizontal component of velocity does not change during the flight, so . The vertical component of velocity is found as
.
9We choose our origin at the initial position of the projectile. After 3.00 s, it is at ground level, so the vertical displacement is .
To find H, we use , which becomes
, or .
10The speed of the car when it reaches the edge of the cliff is
.
Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as it reaches the water is
,
or.
(b)The time of flight is
.
(a)The horizontal displacement of the car during this time is
.
11(a)At the highest point of the trajectory, the projectile is moving horizontally with velocity components of and
.
(b)The horizontal displacement is and, from , the vertical displacement is
.
The straight line distance is
.
12The velocity of a canoe relative to the shore is given by , where is the velocity of the canoe relative to the water and is the velocity of the water relative to shore.
Applied to the canoe moving upstream, this gives
(1)
and for the canoe going downstream
(2)
(a)Adding equations (1) and (2) gives
, so .
(b) Subtracting (1) from (2) yields
, or .
13The time to reach the opposite side is .
When the motorcycle returns to the original level, the vertical displacement is . Using this in the relation gives a second relation between the takeoff speed and the time of flight as:
or .
Substituting the time found earlier into this result yields the required takeoff speed as
.
14For a projectile fired with initial speed at angle above the horizontal, the time required to achieve a horizontal displacement of is . The vertical displacement of the projectile at this time is
Solving for the initial speed gives .
With ,
we find the required initial speed to be .
15We shall first find the initial velocity of the ball thrown vertically upward. At its maximum height, and . Hence, gives
, or .
In order for the second ball to reach the same vertical height as the first, the second must have the same initial vertical velocity. Thus, we find as
.
16The components of the initial velocity are
, and .
The time for the ball to move 10.0 m horizontally is .
At this time, the vertical displacement of the ball must be
.
Thus, becomes
,
which yields
17(a)The components of the vectors are
Vector / x-component (cm) / y-component (cm)/ 0 / 104
/ 46.0 / 19.5
/ 0 / 84.0
/ 38.0 / 20.2
The sums and are computed as:
or .
(b)To normalize, multiply each component in the above calculation by the appropriate scale factor. The scale factor required for the components of is , and the scale factor needed for components of is . After using these scale factors and recomputing the vector sums, the results are:
.
The difference in the normalized vector sums is .
/ 50.9 / 137
- / –45.1 / –124
/ 5.74 / 12.8
Therefore, , and
, or .
18The components of the three displacements are:
Displacement / x-component (paces) / y-component (paces)75.0 paces @ 240° / –37.5 / –65.0
125 paces @ 135° / –88.4 / +88.4
100 paces @ 160° / –94.0 / +34.2
The resultant displacement is then
or.