Question 1

An ant travels 30 cm east, 25 cm north, and finally 15 cm west. What is the direction of its displacement with respect to its starting point?

Accepted Answer

The ant's final displacement in the east-west direction is 30 cm east - 15 cm west = 15 cm east. In the north-south direction, it is 25 cm north. The angle of displacement, θ, can be found using tan(θ) = opposite/adjacent = 25/15. Solving for θ gives an angle of approximately 59º north of east.

Question 2

The interstate on-ramp is 1.5 km west of your home, and on leaving for grandmother's house you are only able to accomplish an average velocity of 12 km/hr between your home and the on-ramp. At the on-ramp, you wait for traffic for 2.0 min before proceeding toward grandmother's house. If you wish to arrive there, which is 15 km south of the on-ramp, over a river and through some woods, in a total elapsed time of 25.0 minutes, what average speed must you maintain on the interstate highway?

Accepted Answer

First, convert all units to consistent measurements: 1.5 km to the on-ramp, 15 km south to grandmother's house, and total time of 25 minutes (or 25/60 hours). The time spent waiting is 2 minutes (or 2/60 hours), and the time to reach the on-ramp at 12 km/hr is 1.5 km / 12 km/hr = 0.125 hours. Subtract the time to reach the on-ramp and the waiting time from the total time to find the time available for the interstate travel: 25/60 - 0.125 - 2/60 = 0.3083 hours. To find the required speed to cover 15 km in 0.3083 hours, divide the distance by the time: 15 km / 0.3083 hours ≈ 48.65 km/hr. However, since none of the provided options match this calculation and the question specifically asks for the average speed needed on the interstate, it's important to reevaluate the calculation steps and assumptions. Given the discrepancy, the correct approach involves recalculating the available time for the 15 km distance correctly and ensuring all conversions and calculations align with the question's requirements. The mistake in the initial explanation was in the calculation and interpretation of the required speed, not accounting for the correct total and available time for the interstate portion of the trip. The correct calculation should consider the total time allowed for the trip, subtract the time taken to reach the on-ramp and the waiting time, and then calculate the necessary speed to cover the remaining distance to grandmother's house in the remaining time.

Question 3

An ant crawls in a straight line at a constant speed of 0.35 m/s, beginning in the corner of a square classroom that is 5.00 m on a side. It first crosses the diagonal of the room and then turns to the right, continuing at 0.45 m/s to the far corner of the same wall from which it began. What was the ant's average speed for the whole trip?

Accepted Answer

To find the average speed, we need to use the formula:
average speed = total distance / total time

First, let's calculate the total distance. The ant first crosses the diagonal of the room, which has a length of:

diagonal = 5.00 m x √2 ≈ 7.07 m

Then, the ant continues to crawl along one of the walls, which has a length of 5.00 m. So, the total distance is:

total distance = diagonal + wall = 7.07 m + 5.00 m = 12.07 m

Next, we need to calculate the total time. For this, we can use the formula:

time = distance / speed

The time taken by the ant to cross the diagonal is:

time1 = diagonal / speed1 = 7.07 m / 0.35 m/s ≈ 20.20 s

The time taken by the ant to crawl along the wall is:

time2 = wall / speed2 = 5.00 m / 0.45 m/s ≈ 11.11 s

So, the total time is:

total time = time1 + time2 ≈ 31.31 s

Finally, we can calculate the average speed as:

average speed = total distance / total time ≈ 12.07 m / 31.31 s ≈ 0.39 m/s

Therefore, the best choice is D.

Question 4

A baseball is struck by a batter at an angle such that it reaches its maximum height at the location of the pitcher-and just grazes the top of the pitcher's head (at an elevation of 1.50 m above the launch point of the baseball). The pitcher is a horizontal distance of 20.0 m from the point at which the ball was struck. What was the launch speed for the baseball?

Accepted Answer

The launch speed can be found using the equations of motion under gravity. Given that the maximum height is 1.50 m and the horizontal distance is 20.0 m, we can use these values to calculate the initial velocity. The time to reach the maximum height is found using $t = \sqrt{\frac{2h}{g}}$, where $h$ is the height and $g$ is the acceleration due to gravity (9.8 m/s$^2$). The total time of flight is double this time since the time to go up is the same as the time to come down. The horizontal velocity ($v_x$) is constant, and using the total time of flight, we can find $v_x = \frac{d}{t}$, where $d$ is the horizontal distance. The initial velocity ($v_0$) can then be calculated knowing the vertical component of the velocity ($v_y$) at launch, which is related to the maximum height. The correct calculation leads to an initial velocity of approximately 36.6 m/s.

Question 5

Vector A has a magnitude of 3.0 units and makes an angle of −90.0º with the positive x-axis. Vector B has a magnitude of 4.0 units and makes an angle of −120º with the positive x-axis. What is the direction of the vector sum A + B relative to the positive x-axis?

Accepted Answer

To find the direction of the vector sum $A + B$, we first need to break each vector into its x and y components and then sum these components. For vector A:- $A_x = 3.0 \cos(-90^\circ) = 0$ (since $\cos(90^\circ) = 0$)- $A_y = 3.0 \sin(-90^\circ) = -3.0$ (since $\sin(-90^\circ) = -1$)For vector B:- $B_x = 4.0 \cos(-120^\circ) = 4.0 \times \cos(120^\circ) = 4.0 \times (-0.5) = -2.0$- $B_y = 4.0 \sin(-120^\circ) = 4.0 \times \sin(120^\circ) = 4.0 \times (\sqrt{3}/2) = 2\sqrt{3}$Summing the components:- $Sum_x = A_x + B_x = 0 - 2.0 = -2.0$- $Sum_y = A_y + B_y = -3.0 + 2\sqrt{3}$The direction $\theta$ of the resultant vector relative to the positive x-axis is given by $\theta = \tan^{-1}\left(\frac{Sum_y}{Sum_x}\right)$.Plugging in the values:- $\theta = \tan^{-1}\left(\frac{-3.0 + 2\sqrt{3}}{-2.0}\right)$This calculation gives a value around -107 degrees, indicating that the vector sum points in a direction of approximately -107º relative to the positive x-axis.

Question 6

A displacement vector D is given as 40.0 m at an angle of 60.0 degrees NORTH of EAST. The D_xcomponent and theD_y component of the vector are

Accepted Answer

Using trigonometry, we can find that the x component of the vector is 40.0 m * cos(60.0 degrees) = 20.0 m and the y component of the vector is 40.0 m * sin(60.0 degrees) = 34.6 m. Therefore, the answer is Choice C.

Question 7

An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square classroom. It then turns exactly 90 degrees to the right and proceeds an additional 4.0 m with a constant speed of 0.44 m/s, reaching the other corner of the same wall from which it began. What was the ant's average speed for the full trip from corner to corner?

Accepted Answer

The answer of An ant crawls in a straight line...

Question 8

The interstate on-ramp is 5.5 km west of your home, and on leaving for grandmother's house you are only able to accomplish an average velocity of 38 km/hr between your home and the on-ramp. At the on-ramp, you wait for traffic for 2.0 min before proceeding toward grandmother's house. If you wish to arrive there, which is 15 km south of the on-ramp, over a river and through some woods, having achieved an average velocity of 42 km/hr for the whole trip, how fast do you need to drive south on the interstate highway?

Accepted Answer

We can use the equation:

average velocity = total distance / total time

Let's calculate the total distance of the trip first:

distance from home to on-ramp = 5.5 km
distance from on-ramp to grandmother's house = 15 km

total distance = 5.5 km + 15 km = 20.5 km

Let's calculate the total time of the trip now:

time from home to on-ramp = distance / velocity = 5.5 km / 38 km/hr = 0.145 hours
time waiting at on-ramp = 2.0 min = 0.033 hours
time from on-ramp to grandmother's house = distance / velocity = 15 km / 42 km/hr = 0.357 hours

total time = 0.145 hours + 0.033 hours + 0.357 hours = 0.535 hours

Now we can use the equation to find the needed average velocity for the whole trip:

average velocity = total distance / total time = 20.5 km / 0.535 hours = 38.3 km/hr

Since the average velocity for the whole trip needs to be 42 km/hr, the required velocity for the interstate highway segment would be:

velocity = (required average velocity * total time - time from home to on-ramp * velocity from home to on-ramp) / time from on-ramp to grandmother's house

velocity = (42 km/hr * 0.535 hours - 0.145 hours * 38 km/hr) / 0.357 hours

velocity = 74 km/hr

Therefore, the answer is C: 74 km/hr.

Question 9

A displacement vector D is given as 40.0 m at an angle of 60.0 degrees EAST of NORTH. The D_x component and the D_y component of the vector are

Accepted Answer

To find the D_x and D_y components of the vector, we need to resolve it into its horizontal (x) and vertical (y) components using trigonometry. D_x = D * sin(60) = 40.0 m * sin(60) = +34.6 m D_y = D * cos(60) = 40.0 m * cos(60) = +20.0 m Therefore, the best choice is A.

Question 10

A displacement vector of magnitude 15.2 km directed along the positive x-axis combines with a second vector to produce a total displacement of magnitude 17.9 km. What is the magnitude of the second vector?

Accepted Answer

The problem does not specify the direction of the second vector relative to the first vector or the x-axis, which is necessary to calculate the magnitude of the second vector using vector addition principles.

Question 11

An ant crawls in a straight line at a constant speed of 0.24 m/s for a distance of 3.0 m, beginning in the corner of a square classroom. It then turns exactly 90 degrees to the right, and proceeds an additional 4.0 m, reaching the far corner of the same wall from which it began. If the second leg of the journey was crawled in half the amount of time as the first, what was the ant's average speed for the whole trip?

Accepted Answer

For the first leg of the journey, the ant crawls a distance of 3.0 m at a speed of 0.24 m/s. Using the formula speed = distance/time, we can solve for the time it takes: 
time = distance/speed = 3.0 m / 0.24 m/s = 12.5 s

For the second leg of the journey, the ant crawls a distance of 4.0 m, which is twice the speed of the first leg since it takes half the time. Using the formula speed = distance/time, we can solve for the new speed: 
time = 12.5 s / 2 = 6.25 s
speed = distance/time = 4.0 m / 6.25 s = 0.64 m/s

To find the average speed for the whole trip, we can use the formula: 
average speed = total distance / total time

The total distance is the sum of the distances for both legs, which is 3.0 m + 4.0 m = 7.0 m.

The total time is the sum of the time for both legs, which is 12.5 s + 6.25 s = 18.75 s.

Putting it all together: 
average speed = total distance / total time = 7.0 m / 18.75 s = 0.37 m/s

Therefore, the answer is C.

Question 12

A beetle runs an erratic pattern on a square kitchen table that is 1.5 m on a side. The sides are arranged along the directions north/south and east/west. It begins at the southwest corner and travels east along one side to the adjacent corner in 20.0 s. It then turns north and reaches the northeast corner 12.5 s later. What is the magnitude of the beetle's average velocity for the whole trip?

Accepted Answer

The beetle covers a total distance of 3 meters (1.5m + 1.5m) in a total time of 32.5 seconds (20s + 12.5s). Therefore, the average velocity is:

Average Velocity = Total Distance ÷ Total Time

Average Velocity = 3m ÷ 32.5s

Average Velocity ≈ 0.065 m/s

Therefore, the answer is choice C, 0.065 m/s.

Question 13

Vector A has a magnitude of 3.0 units and makes an angle of −90.0º with the positive x-axis. Vector B has a magnitude of 4.0 units and makes an angle of −120º with the positive x-axis. What is the magnitude of the vector sum A + B?

Accepted Answer

To find the magnitude of the vector sum $A + B$, we first break down vectors $A$ and $B$ into their components. For vector $A$, since it makes an angle of $-90.0^\circ$ with the positive x-axis, it is directed entirely along the negative y-axis. Thus, its components are $A_x = 0$ and $A_y = -3.0$ units.For vector $B$, with a magnitude of 4.0 units and an angle of $-120^\circ$ from the positive x-axis, we calculate its components using trigonometry. The angle $-120^\circ$ is equivalent to $240^\circ$ in standard position, which places it in the third quadrant where both x and y components are negative. The components are:- $B_x = 4.0 \cos(-120^\circ) = 4.0 \cos(240^\circ) = -2$ units (since $\cos(240^\circ) = -1/2$)- $B_y = 4.0 \sin(-120^\circ) = 4.0 \sin(240^\circ) = -2\sqrt{3}$ units (since $\sin(240^\circ) = -\sqrt{3}/2$)The sum of the vectors $A + B$ in component form is:- $x$-component: $A_x + B_x = 0 - 2 = -2$ units- $y$-component: $A_y + B_y = -3.0 - 2\sqrt{3}$ unitsThe magnitude of the vector sum is calculated using the Pythagorean theorem:$$|A + B| = \sqrt{(-2)^2 + (-3.0 - 2\sqrt{3})^2} = \sqrt{4 + 9 + 12\sqrt{3} + 12} = \sqrt{25 + 12\sqrt{3}}$$Approximating $\sqrt{3} \approx 1.732$, we get:$$|A + B| \approx \sqrt{25 + 12(1.732)} = \sqrt{25 + 20.784} = \sqrt{45.784} \approx 6.8$$ units.

Question 14

You drive five blocks due north, five blocks due east, and another two blocks due north. What is the magnitude of your displacement?

Accepted Answer

The total displacement can be found by considering the northward and eastward movements separately. First, you move 7 blocks north (5 + 2) and 5 blocks east. The displacement is the hypotenuse of a right-angled triangle with sides of 7 and 5 blocks. Using the Pythagorean theorem, $ \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} $, which is approximately 8.6 blocks.

Question 15

A vector of magnitude 3.8 directed along the positive x-axis combines with a second vector directed along the positive y-axis to produce a total vector of magnitude 6.1. What is the magnitude of the second vector?

Accepted Answer

We can use the Pythagorean theorem to solve this problem. Let's call the magnitude of the second vector "y".

We know that the total vector has a magnitude of 6.1, so:

6.1² = 3.8² + y²

Simplifying:

37.21 = 14.44 + y²

Subtracting 14.44 from both sides:

22.77 = y²

Taking the square root of both sides (and remembering that we only care about the positive root, since y is a magnitude):

y ≈ 4.8

So the magnitude of the second vector is approximately 4.8. The answer is D.

Question 16

The y-component of a vector is −15.0, while its magnitude is 27.2. Which could be the vector's direction?

Accepted Answer

The answer of The y-component of a vector is −15.0,...

Quiz 3: Motion in a Plane