Part 1: Problems in Vector Addition

1. A bus is driven 215 km west and then 85 km southwest. What is the displacement of the bus from the point of origin

(magnitude and direction)?

The resultant vector displacement of the car is given by the green vector below. The components are shown in red (215 km west, 85 km southwest).

The westward displacement is 215 + 85 cos 45⁰ = 275.1 km

The south displacement is 85 sin 45⁰ = 60.1 km.

The resultant displacement has a magnitude of (275.1² + 60.1²)^1/2= 281.6 km which is written as 282 km (why?).

The direction (from the starting point - the origin shown above) is given by

q = tan^-1 (60.1/275.1) = 12.3⁰, or 12.3⁰ degrees south of West

2. A delivery truck travels 18 blocks north, 10 blocks east, and 16 blocks south. What is its final displacement?

Displacement: 10.2 blocks

Direction: 11 degrees North of East (NOTE: DID NOT ASK FOR DIRECTION)

The truck has a displacement of 18 + (-16) = 2 blocks north and 10 blocks east. The resultant (shown in red above) has a magnitude of (2² + 10²)^1/2 = 10.2 blocks and a direction of tan^-1(2/10) = 11 degrees North of East.

3. If V_x = 6.80 units and V_y = -7.40 units, determine the magnitude and direction of (Resulting vector).

Given that V_x= 6.80 units and V_y= -7.40 units, the magnitude of is given by V = (V²x + V²y)^1/2 = 10.05 units

The direction is given by an angle of q = tan^-1(-7.40/6.80) = 47 degrees below the positive x axis

4. The components of a vector can be written V_x, V_y, V_z. What are the components and length of a vector which is the

sum of the two vectors, and whose components are (8.0, -3.7, 0.0) and (3.9, -8.1, -4.4)? OMIT THIS PROBLEM

The sum is found by adding the components of the two vectors (8.0, -3.7, 0.0 and (3.9, -8.1, -4.4) to give

(11.0, -11.8, -4.4).

The length is ((11.9)²+ (-11.8)² + (-4.4)²))^1/2 = 17.3

5. Vector is 6.6 units long and points along the negative x axis. Vector is 8.5 units long and points at +45⁰ to the positive x

axis.

a. What are the x and y components of each vector?

V_1x = -6.6 units, V_1y= 0 units

V_2x = 8.5 cos45 = 6.0 units V_2y = 8.5 sin 45 = 6.0 units

b. Determine the sum (magnitude and angle).

We add the components = (-0.6, 6.0)

The length of the resultant vector is given by (-0.6² + 6.0²)^1/2 = 6.0 units

the angle is given by tan^-1(6.0/0.6) = 84 degrees

Part 2: Response with Discussion. Clearly explain the logic behind your answers

1. A football player kicks a ball very high and then runs in a straight line and catches it. Which had the greater displacement, the

football player or the ball?

Since both the kicker and the ball started their motion at the same location (where the ball was hit) and ended their motion at the same location (where the ball was caught), the displacement of both was the same.

2. If = + , is necessarily greater than and/or ?

No.

The magnitude of the vector sum need not be larger than the magnitude of either contributing vector. For example, if the two vectors being added are the exact opposite of each other, the vector sum will have a magnitude of 0.

3. Two vectors have length = 3.5 km and = 4.0 km. What are the maximum and minimum magnitudes of their vector sum?

If the two vectors are in the same direction, the magnitude of their sum will be a maximum, and will be 7.5 km.

If the two vectors are in the opposite direction, the magnitude of their sum will be a minimum, and will be 0.5 km.

If the two vectors are oriented in any other configuration, the magnitude of their sum will be between 0.5 km and 7.5 km.

4. Can the magnitude of a vector ever

(a) be equal to one of its components

Yes

The magnitude of a vector can equal the length of one of its components if the other components of the vector are all 0; i.e. if the vector lies along one of the coordinate axes. If the vector lies entirely along the x axis, for example, the y component of the vector is 0 and the x component of the vector is the length of the vector.

(b) be less than one of its components?

The magnitude of a vector can never be less than one of its components, because each component contributes a positive amount to the overall magnitude, through the Pythagorean relationship. The square root of a sum of squares is never less than the absolute value of any individual term. For example

c = ± (a² + b²)^1/2 where a and b are sides of a right triangle and c is the hypotenuse. It should be clear that c cannot be less than a or b. In order for the relation to hold, the largest value that a can have would be c. b would then have to be 0. The square root of c² cannot be greater than c.

5. Can a particle

a. That has a constant speed be accelerating?

Yes, a particle with constant speed can be accelerating, if its direction is changing. Driving on a curved roadway at constant speed would be an example. This is covered in more detail in the next chapter.

b. What about the case if it has constant velocity?

No. A particle with constant velocity cannot be accelerating – its acceleration must be zero. It has both constant speed and constant direction.

6. A student sitting in an enclosed train car, moving at constant velocity, throws a ball straight up into the air in her reference frame.

(a) Where does the ball land?

The ball lands at the same point from which it was thrown inside the train car – back in the thrower’s hand.

(b) Where does the ball land if the car accelerates

If the car accelerates, the ball will land behind the point from which it was thrown. The thrower reacts to the acceleration, the ball does not. This is also not an inertial reference frame because of the acceleration

7. Two rowers, who can row at the same speed in still water, start across a river at the same time. One of the rowers heads

straight across and is pulled downstream somewhat by the current. The other one heads upstream at an angle so as to arrive

at a point opposite the starting point. Which rower reaches the opposite side first?

The rower heading stright across.

Both rowers need to cover the same "cross river" distance.

The rower with the greatest speed in the "cross river" direction will be the one that reaches the other side first.

The current has no bearing on the problem because the current doesn't help either of the boats move across the river.

Thus the rower heading straight across will reach the other side first. All of his rowing effort has gone into crossing the river.

For the upstream rower, some of his rowing effort goes into battling the current, and so his "cross river" speed will be only a fraction of his rowing speed.

Part 3: Problems in Projectile Motion (neglect air resistance)

Break each problem response into the parts shown below.

a. Diagram

b. Variables that are known along with values

c. Variable(s) whose values need to be found

d. Equation(s) to use

e. Calculations

1. A Jack Russell leaps horizontally (without getting hurt-she has special powers) from a 6.5-m-high rock with a speed of

3.5 m/s. How far from the base of the rock will she land?

Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction,

V_x0= 3.5 m/s and a_x = 0.

In the vertical direction V_y0 = 0, a_y = 9.80 m/s², y₀ = 0

and the final location y = 6.5 m.

The time for the tiger to reach the ground is found from applying Eq. 2-11b to the vertical motion.

y = y₀ + v_y0t + (1/2)a_yt²

Substituting in the values and solving for t gives 1.15 s

The horizontal displacement is calculated from the constant horizontal velocity.

delta x = v_xt = (3.5 m/s)(1.15 sec) = 4.0 m

2. A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below.

a. How high was the cliff

Choose downward to be the positive y direction. The origin will be at the point where the diver dives from the cliff.

In the horizontal direction

v_x0 = 1.8 m/s and a_x = 0.

In the vertical direction

v_y0 = 0, a_y = 9.80 m/s² , y₀ = 0

and the time of flight is t = 3.0 s. The height of the cliff is found by applying equation 2-11b to the vertical motion.

y = y₀ + v_y0t + (1/2)a_yt² gives y = 44 m

b. How far from its base did the diver hit the water?

The distance from the base of the cliff to where he strikes the water is found from the horizontla motion at constant velocity

delta x = vxt = (1.8 m/s)(3 s) = 5.4 m