Question 1

A boat,whose speed in still water is 8.0 m/s,crosses a river with a current of 6.0 m/s.If the boat heads perpendicular to the current,what is the speed of the boat relative to an observer standing on the shore as it crosses the river?

Accepted Answer

The speed of the boat relative to an observer on the shore is the resultant of the boat's speed in still water and the current's speed. This can be found using the Pythagorean theorem, since the boat's direction and the current's direction are perpendicular to each other: $\sqrt{(8.0)^2 + (6.0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.0$ m/s.

Question 2

An airplane flies between two points on the ground that are 500 km apart.The destination is directly north of the point of origin of the flight.The plane flies with an airspeed of 120 m/s.If a constant wind blows at 10 m/s toward the west during the flight,what direction must the plane fly relative to the air to arrive at the destination?

Accepted Answer

To counteract the westward wind and reach the destination directly north, the plane must fly slightly east of north. The angle can be found using trigonometry, specifically the arctangent function: angle = arctan(wind speed / plane speed) = arctan(10 m/s / 120 m/s) ≈ 4.8° east of north.

Question 3

A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s.During this swim,she is swept downstream 480 m.What is the speed of the current?

Accepted Answer

Let's assume the speed of the swimmer in still water is v and the speed of the current is u.
Then, the swimmer's speed while swimming across the river would be (v^2 + u^2)^0.5, and the net velocity relative to the ground would be v - u.
We know that the swimmer covers the 200 m distance in 6 min 40 s (i.e., 400 seconds). So, we can write the following equation:
200 = (v - u) x 400
=> (v - u) = 0.5 m/s    ------ (Equation 1)
Also, we know that during this swim, the swimmer is swept downstream by 480 m. So, we can write the following equation:
480 = (v + u) x 400
=> (v + u) = 1.2 m/s    ------ (Equation 2)
Now, if we subtract Equation 1 from Equation 2, we get:
2u = 0.7 m/s
=> u = 0.35 m/s
Therefore, the speed of the current is 0.35 m/s.

Question 4

A plane has an eastward heading with an airspeed of 156 m/s.A 20.0 m/s wind is blowing southward at the same time as the plane is flying.The velocity of the plane relative to the ground is

Accepted Answer

The velocity of the plane relative to the ground can be found using vector addition. The eastward velocity (156 m/s) and the southward wind velocity (20 m/s) form a right-angled triangle. The magnitude of the resultant velocity (ground speed) is $\sqrt{156^2 + 20^2} = 157$ m/s. The angle $\theta$ south of east is given by $\tan^{-1}(20/156) = 7.31^\circ$.

Question 5

On a calm day with no wind,you can run a 1500-m race at a velocity of 4.0 m/s.If you run the same race on a day when you have a constant headwind that slows your speed by 2.0 m/s,how much time would it take you to finish the race?

Accepted Answer

The answer of On a calm day with no wind,you...

Question 6

A plane has an airspeed of 142 m/s.A 30.0 m/s wind is blowing southward at the same time as the plane is flying.What must be the heading of the plane in order to move directly eastward relative to the ground?

Accepted Answer

The correct heading is 12.2° north of east. This is because the plane must fly into the wind in order to move directly eastward relative to the ground. The wind is blowing southward, so the plane must fly northward in order to counteract the wind's effect. The plane's airspeed is 142 m/s, and the wind speed is 30.0 m/s, so the plane must fly at an angle of arctan(30.0/142) = 12.2° north of east in order to move directly eastward relative to the ground.

Question 7

A plane has an airspeed of 142 m/s.A 16 m/s wind is blowing southward at the same time as the plane is flying.If the velocity of the plane relative to the ground is directly eastward,what is the magnitude of that velocity?

Accepted Answer

We need to use vector addition since we have two velocities at different directions. Let eastward direction be positive and northward direction be negative.

Let's call the velocity of the plane relative to the ground as V and the velocity of the wind as W. We know that V has a component going eastward and a component going southward. We also know that W is going southward.

Using Pythagorean theorem, we can find the magnitude of V:

V² = (142 m/s)² + (16 m/s)² 
V² = 20160 
V = √20160 
V ≈ 141 m/s

Therefore, the magnitude of the velocity of the plane relative to the ground is 141 m/s.

Question 8

A plane flies directly from city A to city B,which are separated by 2300 mi.From A to B,the plane flies into a 65 mi/h headwind.On the return trip from B to A,the wind velocity is unchanged.The trip from B to A takes 
Less than the trip from A to B.What is the airspeed (assumed constant)of the plane relative to the air (that is,the airspeed)?

Accepted Answer

The answer of A plane flies directly from city A...

Question 9

A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s.During this swim,she is swept downstream 480 m.How fast can she swim in still water?

Accepted Answer

The answer of A swimmer heading directly across a river...

Quiz 3: Vectors in Physics