Question 1

A river flows toward the East at 1.7 m/s.Steve can swim 2.0 m/s in still water.If he swims,relative to the water at a direction of 33 degrees west of north,how long will it take him to cross the 75 m wide river?
A) 45 s
B) 39 s
C) 123 s
D) 27 s

Accepted Answer

The correct answer is A.Explanation:The velocity of the river is 1.7 m/s toward the East. Steve's velocity relative to the water is 2.0 m/s at a direction of 33 degrees west of north. We can use the law of cosines to find the magnitude of Steve's velocity relative to the ground:v = sqrt(v_water^2 + v_river^2 - 2v_water*v_river*cos(theta))v = sqrt(2.0^2 + 1.7^2 - 2*2.0*1.7*cos(33))v = 2.7 m/sNow we can use the formula distance = velocity * time to find the time it takes Steve to cross the river:t = d/vt = 75 m / 2.7 m/st = 27.8 sTherefore, the correct answer is A) 45 s.

Question 2

A model airplane is flying north at 12.5 m/s initially - and 25 seconds later is observed heading 30 degrees west of north at 25 m/s.What is the magnitude of the average acceleration during this time interval? A) 0.37 m/s² B) 0.50 m/s² C) 0.62 m/s² D) 0.87 m/s²

Accepted Answer

First, we need to find the change in velocity of the airplane:
- The initial velocity is 12.5 m/s north
- 25 seconds later, the velocity is 25 m/s at 30 degrees west of north.
- To find the change in velocity, we need to find the vector sum of the two velocities.

Using trigonometry, we can break down the second velocity vector into its x and y components:
- The y component is (25 m/s)Sin(30) = 12.5 m/s
- The x component is (25 m/s)Cos(30) = 21.7 m/s

To find the vector sum, we add the x and y components separately:
- The final x component is 12.5 m/s (north) + 21.7 m/s (west) = 9.2 m/s (west)
- The final y component is 25 m/s (north) - 12.5 m/s (north) = 12.5 m/s (north)

Using the Pythagorean theorem, we can find the magnitude of the final velocity:
Vf = sqrt((9.2 m/s)^2 + (12.5 m/s)^2) = 15.2 m/s

Now we can find the average acceleration:
a = (Vf - Vi) / t
- Vi is the initial velocity, which is 12.5 m/s north
- t is the time interval, which is 25 seconds
a = (15.2 m/s - 12.5 m/s north) / 25 s = 0.62 m/s^2 (southwest)

Note that the direction of the acceleration is southwest because the change in velocity is partly in the west direction and partly in the south direction. However, the question only asks for the magnitude of the acceleration, so we do not need to provide the direction in the answer.

Question 3

A river flows toward the East at 2.2 m/s.Steve can swim 2.5 m/s in still water.If he wishes to cross the river so that his direction is northward,how long will it take him to cross the 75 m wide river?
A) 34 s
B) 39 s
C) 30 s
D) 63 s

Accepted Answer

The answer of A river flows toward the East at...

Question 4

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.If he begins at the Southwest corner,and travels along one side to the adjacent corner in 20.0 s.He then turns north,and reaches the Northeast corner 12.5 s later.What is the magnitude of his average velocity for the whole trip?
A) 0.092 m/s
B) 0.098 m/s
C) 0.065 m/s
D) 0.033 m/s

Accepted Answer

The beetle travels a total distance of 3.0 m (1.5 m east and 1.5 m north) in a total time of 32.5 s. The average velocity is the total displacement divided by the total time, which is 3.0 m / 32.5 s = 0.092 m/s.

Question 5

A 450 g baseball is pitched,reaching the batter with a speed of 37 m/s and is hit such that it returns on the same line at 48 m/s.If the ball was in contact with the bat for 0.22 seconds,what was the average net force experienced by the ball during the time of contact?
A) 170 N
B) 98 N
C) 77 N
D) 18 N

Accepted Answer

The change in velocity is 48 m/s - (-37 m/s) = 85 m/s. The mass of the baseball is 0.45 kg. Using the formula for force, F = m * Δv / Δt, we get F = 0.45 kg * 85 m/s / 0.22 s = 170 N.

Question 6

A river flows toward the East at 1.5 m/s.Steve can swim 2.2 m/s in still water.If he wishes to cross the river so that his direction is northward,what direction relative to the water must he swim?
A) 55$^{\circ}$ north of west
B) 43$^{\circ}$ west of north
C) 34$^{\circ}$ north of west
D) 47$^{\circ}$ west of north

Accepted Answer

The answer of A river flows toward the East at...

Question 7

An object travels 7.5 m/s toward the West and under the influence of a constant net force of 5.2 kN,4.3 s later,it is travelling 2.5 m/s toward the East.What is its mass?
A) 2200 kg
B) 4500 kg
C) 6000 kg
D) 12000 kg

Accepted Answer

The change in velocity is 10 m/s (from 7.5 m/s West to 2.5 m/s East). Using the formula F = ma, where F is the net force (5200 N), m is the mass, and a is the acceleration (change in velocity/time), we find m = F * t / Δv = 5200 N * 4.3 s / 10 m/s = 2236 kg, which is closest to 2200 kg.

Question 8

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 average speed for the full trip from corner to corner?
A) 0.34 m/s
B) 0.30 m/s
C) 0.21 m/s
D) 0.23 m/s
E) 0.32 m/s

Accepted Answer

The total distance traveled is 3.0 m + 4.0 m = 7.0 m. The time for the first segment is 3.0 m / 0.24 m/s = 12.5 s, and for the second segment is 4.0 m / 0.44 m/s = 9.09 s. The total time is 12.5 s + 9.09 s = 21.59 s. The average speed is total distance / total time = 7.0 m / 21.59 s ≈ 0.32 m/s.

Question 9

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 4.00 m,reaching the far corner of the same wall from which it began.If the second leg was also crawled in half the amount of time as required for the first leg,what was the ant's average speed for the whole trip?
A) 0.80 m/s
B) 0.27 m/s
C) 0.37 m/s
D) 0.44 m/s

Accepted Answer

First, we need to find the time it took the ant to crawl the first leg:
time = distance / speed
time = 3.0 m / 0.24 m/s = 12.5 s

Next, we know that the second leg took half the time of the first leg, so it took:
6.25 s

Now we can find the total distance travelled by the ant:
total distance = 3.0 m + 4.0 m = 7.0 m

And we can find the total time taken by the ant:
total time = 12.5 s + 6.25 s = 18.75 s

Finally, we can find the average speed of the ant for the entire trip:
average speed = total distance / total time
average speed = 7.0 m / 18.75 s = 0.37 m/s

Therefore, the best choice is C, 0.37 m/s.

Question 10

You drive your car 5.0 km due East at 35 km/hr,and suddenly realize that you forgot your wallet.So,you return home,driving West at 40 km/hr,and upon arrival spend 10 minutes looking for it,and finally go back on the road for a total of 57.0 km due East.If your average velocity was 40 km/hr for the whole journey,what was your average speed on your last leg.
A) 49 km/hr
B) 40 km/hr
C) 45 km/hr
D) 56 km/hr

Accepted Answer

Average speed = Total distance/Total timeTotal distance = 5.0 km + 5.0 km + 57.0 km = 67.0 kmTotal time = 5.0 km/35 km/hr + 5.0 km/40 km/hr + 10 min + 57.0 km/40 km/hr= 0.143 hr + 0.125 hr + 0.167 hr + 1.425 hr= 1.86 hrAverage speed = 67.0 km/1.86 hr = 36.0 km/hr

Question 11

An object travels 7.5 m/s toward the West and under the influence of a constant net force of 5.2 kN,it comes to rest in 3.2 s.What is its mass?
A) 1600 kg
B) 12000 kg
C) 1100 kg
D) 690 kg

Accepted Answer

We can use the equation Fnet = ma, where Fnet is the net force, m is the mass, and a is the acceleration.
Since the object comes to rest, its final velocity is 0 m/s. Therefore, we can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
Thus, 0 m/s = 7.5 m/s + a(3.2 s) 
Solving for the acceleration, we get a = -2.34375 m/s^2 (negative sign indicates that the acceleration is in the opposite direction of the initial velocity).
Plugging this value into Fnet = ma, we get 5.2 kN = m(-2.34375 m/s^2).
Converting kN to N and solving for m, we get m ≈ 1101 kg.
Therefore, the best choice is C.

Question 12

The velocity of an airplane with respect to the ground is 2.00 $\times$ 10² m/s at an angle of 30.0 degrees NORTH of EAST.The velocity of the airplane with respect to the air is 1.50 $\times$ 10² m/s at an angle of 60.0 degrees NORTH of EAST.What is the velocity of the air with respect to the ground? A) 112 m/s at 50.0 degrees NORTH of EAST B) 102 m/s at 17 degrees SOUTH of EAST C) 134 m/s at 49 degrees SOUTH of EAST D) 134 m/s at 17 degrees NORTH of EAST E) 102 m/s at 65 degrees NORTH of EAST

Accepted Answer

To find the velocity of the air with respect to the ground, subtract the velocity of the airplane with respect to the air from the velocity of the airplane with respect to the ground using vector subtraction. The correct result is 102 m/s at 17 degrees SOUTH of EAST.

Question 13

A baseball is pitched,reaching the batter with a speed of 37 m/s and is hit such that it returns on the same line at 48 m/s.If the ball was in contact with the bat for 0.22 seconds,what was the average acceleration experienced by the ball during the time of contact? A) 220 m/s² B) 41 m/s² C) 390 m/s² D) 170 m/s²

Accepted Answer

The average acceleration is calculated using the formula: a = (v_f - v_i) / t. Here, v_f = 48 m/s (final velocity), v_i = -37 m/s (initial velocity, negative because it's in the opposite direction), and t = 0.22 s. So, a = (48 - (-37)) / 0.22 = 390 m/s².

Question 14

A 78.0 kg object is travelling due North at 15.2 m/s,and experiences a constant net force of 1.72 kN toward the South.How long will it take to stop under the influence of this force?
A) 1.45 s
B) 2.98 s
C) 0.335 s
D) 0.689 s

Accepted Answer

The answer of A 78.0 kg object is travelling due...

Question 15

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?
A) 0.40 m/s
B) 0.16 m/s
C) 0.24 m/s
D) 0.39 m/s

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 16

A 1.7 kg model airplane is flying north at 12.5 m/s initially - and 25 seconds later is observed heading 30 degrees west of north at 25 m/s.What is the magnitude of the average net force on the airplane during this time interval?
A) 0.85 N
B) 0.63 N
C) 1.1 N
D) 1.5 N

Accepted Answer

Firstly, let's find the change in velocity of the airplane during the time interval:

Change in velocity (Δv) = Final velocity - Initial velocity
Δv = 25 m/s at 30 degrees west of north - 12.5 m/s north
Δv = (25*cos(30) - 12.5) i + (25*sin(30)) j
Δv = 18.66 i + 12.5 j  (where i and j are unit vectors in x and y directions respectively)

The time interval is given as 25 seconds.

Now, we can use the equation F = m*a, where a is the acceleration.

Acceleration (a) = Δv / Δt
a = Δv / 25
a = (18.66/25) i + (12.5/25) j
a = 0.7464 i + 0.5 j

Magnitude of acceleration (|a|) = sqrt((0.7464)^2 + (0.5)^2)
|a| = 0.895 m/s^2

Using F = m*a, where m is the mass of the airplane:

Magnitude of net force (|F|) = m*|a|
|F| = 1.7 kg * 0.895 m/s^2
|F| = 1.52 N

Therefore, the closest option is C) 1.1 N.

Question 17

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 - and then are able to proceed 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,how fast do you need to drive south on the interstate highway?
A) 58 km/hr
B) 48 km/hr
C) 78 km/hr
D) 39 km/hr

Accepted Answer

The answer of The interstate on-ramp is 5.5 km West...

Question 18

The velocity of the air with respect to the ground is 50.0 m/s at an angle of 30.0 degrees NORTH of EAST.The velocity of the airplane with respect to the air is 150 m/s at an angle of 60.0 degrees NORTH of EAST.What is the velocity of the airplane with respect to the ground?
A) 158 m/s at 16 degrees NORTH of EAST
B) 103 m/s at 29 degrees SOUTH of EAST
C) 200 m/s at 30 degrees NORTH of EAST
D) 135 m/s at 29 degrees SOUTH of EAST
E) 200 m/s at 30 degrees SOUTH of EAST

Accepted Answer

The velocity of the airplane with respect to the ground is the vector sum of the velocity of the airplane with respect to the air and the velocity of the air with respect to the ground. Calculating this gives a resultant velocity of 200 m/s at 30 degrees NORTH of EAST.

Question 19

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 onramp,you wait for traffic,for 2.0 min - and then are able to proceed toward grandmother's house.If you wish to arrive there,which is 15 km South of the onramp,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?
A) 51 km/hr
B) 67 km/hr
C) 36 km/hr
D) none of these.

Accepted Answer

First, let's convert the total elapsed time of 25.0 minutes to hours:

25.0 minutes = 25.0/60 hours = 0.417 hours

Now, let's find the total distance between the on-ramp and grandmother's house:

1.5 km (distance from home to on-ramp) + 15 km (distance from on-ramp to grandmother's house) = 16.5 km

We know that the time spent at the on-ramp waiting for traffic is 2.0 minutes, which is 2.0/60 hours = 0.033 hours.

So, the time spent driving on the interstate is:

0.417 hours (total elapsed time) - 0.033 hours (time spent at on-ramp) = 0.384 hours

Now we can use the formula:

average speed = total distance / total time

average speed = 16.5 km / 0.384 hours

average speed ≈ 43 km/hr

However, we need to remember that the average velocity from home to the on-ramp was only 12 km/hr. This means that for the first part of the trip, the average speed was lower than the overall average speed. So, to maintain an overall average speed of 43 km/hr, we need to travel faster than 43 km/hr on the interstate.

Let's call the speed we need to maintain on the interstate "x". We can set up the following equation:

(1.5 km / 12 km/hr) + (15 km / x) = 0.384 hours

Solving for x:

1.5/12 + 15/x = 0.384

15/x = 0.384 - 1.5/12

15/x = 0.32

x = 15/0.32 ≈ 47 km/hr

Therefore, we need to maintain an average speed of approximately 47 km/hr on the interstate in order to arrive at grandmother's house in a total elapsed time of 25.0 minutes.

The closest answer choice is B) 67 km/hr, which is not exactly correct, but is the best choice given the options provided.

Quiz 3: Acceleration and Newtons Second Law of Motion

A river flows toward the East at 1.7 m/s.Steve can swim 2.0 m/s in still water.If he swims,relative to the water at a direction of 33 degrees west of north,how long will it take him to cross the 75 m wide river?

A model airplane is flying north at 12.5 m/s initially - and 25 seconds later is observed heading 30 degrees west of north at 25 m/s.What is the magnitude of the average acceleration during this time interval?

A river flows toward the East at 2.2 m/s.Steve can swim 2.5 m/s in still water.If he wishes to cross the river so that his direction is northward,how long will it take him to cross the 75 m wide river?

A 450 g baseball is pitched,reaching the batter with a speed of 37 m/s and is hit such that it returns on the same line at 48 m/s.If the ball was in contact with the bat for 0.22 seconds,what was the average net force experienced by the ball during the time of contact?

A river flows toward the East at 1.5 m/s.Steve can swim 2.2 m/s in still water.If he wishes to cross the river so that his direction is northward,what direction relative to the water must he swim?

An object travels 7.5 m/s toward the West and under the influence of a constant net force of 5.2 kN,4.3 s later,it is travelling 2.5 m/s toward the East.What is its mass?

An object travels 7.5 m/s toward the West and under the influence of a constant net force of 5.2 kN,it comes to rest in 3.2 s.What is its mass?

A baseball is pitched,reaching the batter with a speed of 37 m/s and is hit such that it returns on the same line at 48 m/s.If the ball was in contact with the bat for 0.22 seconds,what was the average acceleration experienced by the ball during the time of contact?

A 78.0 kg object is travelling due North at 15.2 m/s,and experiences a constant net force of 1.72 kN toward the South.How long will it take to stop under the influence of this force?

A 1.7 kg model airplane is flying north at 12.5 m/s initially - and 25 seconds later is observed heading 30 degrees west of north at 25 m/s.What is the magnitude of the average net force on the airplane during this time interval?

The velocity of the air with respect to the ground is 50.0 m/s at an angle of 30.0 degrees NORTH of EAST.The velocity of the airplane with respect to the air is 150 m/s at an angle of 60.0 degrees NORTH of EAST.What is the velocity of the airplane with respect to the ground?