Question 1

The displacement of a particle is _______ the distance the object has traveled.
A) larger than
B) smaller than
C) either larger or smaller than
D) the same as
E) smaller than or equal to

Accepted Answer

Displacement is a vector quantity that refers to the change in position of an object. It can be zero or less than the actual path length (distance) traveled by the object, especially if the object moves back towards its starting point. Therefore, displacement is always smaller than or equal to the distance traveled.

Question 2

A vector has an x component of -5.51 units and a y component of +9.52 units. The angle between the positive direction of the vector and the positive direction of the x axis is
A) 125º
B) 60º
C) 120º
D) 150º
E) -60º

Accepted Answer

The angle θ between the vector and the positive x-axis can be found using the tangent function: tan(θ) = y/x. Here, y = 9.52 and x = -5.51. Calculating the arctan(9.52 / -5.51) gives an angle in the second quadrant (since the x-component is negative and the y-component is positive), which corresponds to 120º.

Question 3

A river is 0.76 km wide. The banks are straight and parallel. The current is 5.0 km/h and is parallel to the banks. A boat has a maximum speed of 3 km/h in still water. The pilot of the boat wishes to travel on a straight line from A to B, where AB is perpendicular to the banks. The pilot should
A) head directly across the river.
B) head 68º upstream from the line AB.
C) head 22º upstream from the line AB.
D) give up. The trip from A to B is not possible with this boat.
E) do none of these.

Accepted Answer

D

Question 4

A vector has an x component of 5.51 units and a y component of -9.52 units. The angle between the positive direction of the vector and the positive direction of the x axis is approximately
A) 30º
B) -30º
C) 60º
D) -60º
E) None of these is correct.

Accepted Answer

The angle between the positive direction of the vector and the positive direction of the x-axis can be found using the formula:
θ = tan⁻¹(y/x)
θ = tan⁻¹(-9.52/5.51)
θ ≈ -59.57º
Since the angle is in the fourth quadrant, we add 360º to get the angle between 0º and 360º:
θ ≈ 300.43º
We want the acute angle, so we subtract from 360º:
θ ≈ 59.57º
Therefore, the angle between the positive direction of the vector and the positive direction of the x-axis is approximately -60º.

Question 5

The position vector of an object is given by  and 4 s later, its position vector is  . The units are in m. The average velocity vector,  in m/s is
A) 
B) 
C) 
D) 
E) None of the above

Accepted Answer

The answer of The position vector of an object is...

Question 6

You walk 12 km south and then 5 km east. Your resultant displacement is
A) 13 km at an angle of 22.6º south of east.
B) 13 km at an angle of 67.4º south of east.
C) 17 km at an angle of 22.6º south of east.
D) 17 km at an angle of 67.4º south of east.
E) None of these is correct.

Accepted Answer

The resultant displacement can be found by drawing a right-angled triangle with one side of length 12 km and the other side of length 5 km. The hypotenuse of this triangle represents the resultant displacement.

Using the Pythagorean theorem, we find that the length of the hypotenuse is:

√(12² + 5²) = √(144 + 25) = √169 = 13 km

To find the angle that the resultant displacement makes with the south direction, we use the tangent function:

tanθ = opposite/adjacent = 5/12
θ = tan⁻¹(5/12) = 22.6º

However, we want to find the angle that the resultant displacement makes with the east direction, which is 90º - θ:

90º - 22.6º = 67.4º

So the resultant displacement is 13 km at an angle of 67.4º south of east, which is choice B.

Question 7

A car is traveling east at 50 km/h. It rounds a curve, and 5 s later it is traveling north at 50 km/h. The magnitude of the average acceleration of the car is
A) zero
B) 20 km/h · s
C) 10 km/h · s
D) 50 km/h · s
E) None of these is correct.

Accepted Answer

The answer of A car is traveling east at 50...

Question 8

A particle has an initial velocity of 4.8 m/s toward the south and a final velocity of 7.1 m/s toward the east. The particle was subject to a constant acceleration for 0.25 s. The magnitude and the direction of the acceleration must have been A) 8.6 m/s² at 34.1º north of east B) 260 m/s² at 34.1º south of east C) 34 m/s² at 34.1º north of east D) 34 m/s² at 34.1º south of east E) 8.6 m/s² at 34.1º south of east

Accepted Answer

We can use the kinematic equations to solve for the acceleration. Let's use the equation: 
v = u + at 
where u is the initial velocity, v is the final velocity, a is the acceleration, and t is the time taken.

Breaking down the initial and final velocities into their components, we have: 
u = -4.8 m/s (toward the south) 
v = 7.1 m/s (toward the east)

Since the acceleration is constant, we can ignore the direction for now and focus on the magnitudes of the velocities: 
|u| = 4.8 m/s 
|v| = 7.1 m/s

Note that we take the absolute value since we only care about the magnitudes.

Now, let's use the given time: 
t = 0.25 s

Substituting in the values we have:

7.1 = -4.8 + a(0.25) 
a = 34 m/s^2

So the magnitude of the acceleration is 34 m/s^2.

To find the direction, we can use trigonometry. We can draw a right triangle where the horizontal leg is the final velocity (7.1 m/s toward the east), the vertical leg is the negative of the initial velocity (-4.8 m/s toward the south), and the hypotenuse is the resultant velocity (which we don't need right now).

The angle we want is the angle between the hypotenuse and the horizontal leg, or equivalently, the angle between the initial and final velocities. We can use the inverse tangent function:

θ = tan^-1(4.8/7.1) 
θ = 34.1º

Note that we take the inverse tangent of the absolute value of the ratio to get the correct quadrant.

So the direction of the acceleration is 34.1º north of east.

Therefore, the best choice is C) 34 m/s^2 at 34.1º north of east.

Question 9

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the velocity vector at P₁ and P₂? A) (1) B) (2) C) (3) D) (4) E) None of these is correct.

Accepted Answer

At P₁, the velocity vector is horizontal and in the positive x-direction, so it is represented by vector (2). At P₂, the velocity vector is vertical and in the negative y-direction, so it is represented by vector (2) as well (but with the arrow pointing downwards). The other vectors do not accurately represent the direction and magnitude of the velocity vector at these points.

Question 10

The position vector of an object is given by  and 4 s later, its position vector is  . The units are in m. The change in the position vector  , in m, is
A) 
B) 
C) 
D) 
E) None of the above

Accepted Answer

The answer of The position vector of an object is...

Question 11

A car is traveling south at 30 km/h. It rounds a curve, and 6 s later it is traveling west at 30 km/h. The magnitude of the average acceleration of the car is
A) zero
B) 60 km/h · s
C) 5 km/h · s
D) 50 km/h · s
E) 7.1 km/h · s

Accepted Answer

Let's assume the initial velocity towards south as u and final velocity towards west as v. Let the average acceleration be a.

We know that acceleration (a) is given by:

a = (v-u)/t

where t is the time taken for the change in velocity.

In this question, time taken (t) is given as 6 seconds.

Hence, we can write:

a = (v-u)/t
  = (30 km/h - (-30 km/h))/6 s
  = 60 km/h/6 s
  = 10 km/h/s

Now, we need to find the magnitude of average acceleration. Magnitude is always positive, hence we can take the absolute value of acceleration.

Magnitude of acceleration = |a| = 10 km/h/s

But, the answer choices are not in km/h/s. We need to convert km/h/s to km/h^2

1 km/h = 1/3600 km/s (since 1 hour = 3600 seconds)

Hence,

10 km/h/s = 10/(3600) km/s^2 = 1/360 km/s^2

1 km/h^2 = 1 km/(h^2)

Hence,

1/360 km/s^2 = (1/360) x (3600/1) km/h^2
             = 10 km/h^2

Therefore, the magnitude of average acceleration of the car is 10 km/h^2, which is approximately equal to 7.1 km/h^2.

Hence, the best choice is E.

Question 12

The position vector locating the point P(5,12) relative to the origin is
A) 12  + 5 
B) -5  - 12 
C) 5  + 12  
D) 13  + 13 
E) 13  - 13

Accepted Answer

The position vector of a point is given by the ordered pair (x,y). In this case, the point P(5,12) has a position vector of (5,12) relative to the origin. Choice C is the only option that matches the ordered pair (5,12) in the position vector format, with the first component being the x-coordinate and the second component being the y-coordinate.

Question 13

The figure shows the position of the moon at two times, about 7 days apart. Which vector best represents the change in the moon's velocity in this time interval?
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of  The figure shows the position of...

Question 14

The initial path of a rocket is 30º above the horizontal, and the horizontal component of the rocket's velocity is 326 m/s. The initial vertical component of the rocket's velocity is
A) 188 m/s
B) 330 m/s
C) 380 m/s
D) 280 m/s
E) 250 m/s

Accepted Answer

We can use trigonometry to find the initial vertical component of the rocket's velocity.

The horizontal component is given as 326 m/s.

Let's call the initial vertical component V.

We can use the trigonometric function tangent to relate the angle of 30º and the ratio of the opposite (V) to the adjacent (326):

tan(30º) = V/326

Solving for V, we get:

V = 326*tan(30º) = 326*(1/√3) = 188 m/s

Therefore, the initial vertical component of the rocket's velocity is 188 m/s.

Question 15

You walk 5 km north and then 12 km east. Your resultant displacement is
A) 13 km at an angle of 22.6º north of east.
B) 13 km at an angle of 67.4º north of east.
C) 17 km at an angle of 22.6º north of east.
D) 17 km at an angle of 67.4º north of east.
E) None of these is correct.

Accepted Answer

The resultant displacement can be found using the Pythagorean theorem, where the total displacement is the hypotenuse of a right triangle with sides of 5 km and 12 km. This gives a displacement of 13 km. The angle can be found using the tangent function: tan(θ) = opposite/adjacent = 5/12. Solving for θ gives an angle of 22.6º north of east.

Question 16

The position vector of an object is given by  and 4 s later, its position vector is  . The units are in m. The magnitude of the change in the position vector  is
A) 0 m
B)  m
C)  m
D) 10 m
E)  m

Accepted Answer

To find the magnitude of the change in position vector, we need to take the difference between the two position vectors given:
11eaa4be_83b6_869d_9e4a_0b3bbbfe3049_TB6080_11 - 11eaa4be_83b6_5f8c_9e4a_154589a8cf37_TB6080_11 = 11eaa4be_83b6_adb0_9e4a_b153cd9ab97e_TB6080_11
The magnitude of this vector is equivalent to its length, which is approximately 10. The correct answer choice is C.

Question 17

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the change in position vector from P₁ to P₂? A) (1) B) (2) C) (3) D) (4) E) None of these is correct.

Accepted Answer

The change in position vector from PS1U1B11S1U1B0 to PS1U1B12S1U1B0 is represented by the vector that points from point S1U1B11 to point S1U1B12. Vector diagram (4) is the only one that accurately shows this change in position vector, as it points in the correct direction and has the same magnitude as the constant speed of the object.

Question 18

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the change in velocity vector from P₁ to P₂? A) (1) B) (2) C) (3) D) (4) E) None of these is correct.

Accepted Answer

Since the object moves with constant speed, its acceleration is zero. Therefore, the change in velocity vector from PS1U1B11S1U1B0 to PS1U1B12S1U1B0 will be in the direction of the tangent to the curve at point S1U1B11. Vector diagram (3) represents this change in velocity vector. In contrast, vector diagrams (1), (2), and (4) do not accurately depict the change in velocity vector.

Question 19

The position vector locating the point P(12,-5) relative to the origin is
A) 12  + 5 
B) -5  - 12 
C) 5  + 12 
D) 12  + 5 
E) 13  - 13

Accepted Answer

The answer of The position vector locating the point P(12,-5)...

Question 20

If an object is moving west, its acceleration
A) is north.
B) is east.
C) is west.
D) is south.
E) may be any direction.

Accepted Answer

The question does not provide enough information about the acceleration of the object. The direction of the acceleration depends on the net force acting on the object. Therefore, the acceleration could be in any direction.

Quiz 3: Motion in Two and Three Dimensions

The displacement of a particle is _______ the distance the object has traveled.

A vector has an x component of -5.51 units and a y component of +9.52 units. The angle between the positive direction of the vector and the positive direction of the x axis is

A river is 0.76 km wide. The banks are straight and parallel. The current is 5.0 km/h and is parallel to the banks. A boat has a maximum speed of 3 km/h in still water. The pilot of the boat wishes to travel on a straight line from A to B, where AB is perpendicular to the banks. The pilot should

A vector has an x component of 5.51 units and a y component of -9.52 units. The angle between the positive direction of the vector and the positive direction of the x axis is approximately

The position vector of an object is given by and 4 s later, its position vector is . The units are in m. The average velocity vector, in m/s is

You walk 12 km south and then 5 km east. Your resultant displacement is

A car is traveling east at 50 km/h. It rounds a curve, and 5 s later it is traveling north at 50 km/h. The magnitude of the average acceleration of the car is

A particle has an initial velocity of 4.8 m/s toward the south and a final velocity of 7.1 m/s toward the east. The particle was subject to a constant acceleration for 0.25 s. The magnitude and the direction of the acceleration must have been

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the velocity vector at P1 and P2?

The position vector of an object is given by and 4 s later, its position vector is . The units are in m. The change in the position vector , in m, is

A car is traveling south at 30 km/h. It rounds a curve, and 6 s later it is traveling west at 30 km/h. The magnitude of the average acceleration of the car is

The position vector locating the point P(5,12) relative to the origin is

The figure shows the position of the moon at two times, about 7 days apart. Which vector best represents the change in the moon's velocity in this time interval?

The initial path of a rocket is 30º above the horizontal, and the horizontal component of the rocket's velocity is 326 m/s. The initial vertical component of the rocket's velocity is

You walk 5 km north and then 12 km east. Your resultant displacement is

The position vector of an object is given by and 4 s later, its position vector is . The units are in m. The magnitude of the change in the position vector is

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the change in position vector from P1 to P2?

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the change in velocity vector from P1 to P2?

The position vector locating the point P(12,-5) relative to the origin is

If an object is moving west, its acceleration

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the velocity vector at P₁ and P₂?

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the change in position vector from P₁ to P₂?

Use the diagram to answer question. The curve shows the position of an object with time. The object moves with a constant speed, v. -Which of the vector diagrams below best represent the change in velocity vector from P₁ to P₂?