3. A private plane flies for 2.0 hours at 150 mph on a bearing of 060°. Then it turns and continues another 3.0 hours at the same speed, but on a bearing of 150°. At the end of this time, how far is the plane from its starting point



Answer :

Answer:

To solve this problem, we'll break it down into two components: the horizontal and vertical components of the plane's motion.

First, let's consider the horizontal motion:

For the first leg of the journey (at a bearing of 060°), the horizontal displacement (

1

d

1

) can be calculated using:

1

=

speed

×

time

d

1

=speed×time

1

=

150

mph

×

2.0

hours

=

300

miles

d

1

=150mph×2.0hours=300miles

For the second leg of the journey (at a bearing of 150°), the horizontal displacement (

2

d

2

) can also be calculated similarly:

2

=

speed

×

time

d

2

=speed×time

2

=

150

mph

×

3.0

hours

=

450

miles

d

2

=150mph×3.0hours=450miles

Now, let's consider the vertical motion:

Since the plane flies at a constant speed, there is no change in altitude, so the vertical displacement is zero.

Now, we'll use the Pythagorean theorem to find the total distance from the starting point, which is the hypotenuse of the right triangle formed by the horizontal and vertical displacements:

Total distance

=

1

2

+

2

2

Total distance=

d

1

2

+d

2

2

Total distance

=

30

0

2

+

45

0

2

Total distance=

300

2

+450

2

Total distance

=

90000

+

202500

Total distance=

90000+202500

Total distance

=

292500

Total distance=

292500

Total distance

540.55

miles

Total distance≈540.55miles

So, at the end of this time, the plane is approximately

540.55

540.55 miles from its starting point.

Step-by-step explanation: