a plane takes off at 10.00am from an airfield, and flies at 120km/h on a bearing of N35W. A second plane takes off at 10.05am from the same airfield, and flies on a bearing of S80E at a speed of 90km/h. How far apart are the planes at 10.25am?
Did you make a sketch?
At the time of our calculation ,
the first plane flew for 25 min or 5/12 hr
the 2nd plance flew for 20 min or 1/3 hr
distance covered by 1st plane = (5/12)(120) = 50 km
distance covered by 2nd plane = (1/3)(90) = 30 km
I found the angle between them to be 135°
looks like a standard cosine law question....
x^2 = 50^2 + 30^2 - 2(50)(30)cos135°
= .....
careful, the cos135° is negative , making that last term positive.
A plane flies in a direction of N70°E for 80 km and then on a bearing of S10°W for 150 km.
a How far is the plane from its starting point?
b What direction is the plane from its starting point?
To find the distance between the two planes at 10:25 am, we need to calculate their respective positions at that time.
First, let's calculate the displacement of the first plane from 10:00 am to 10:25 am.
Time elapsed: 10:25 am - 10:00 am = 25 minutes = 25/60 hours
Distance traveled by the first plane: Speed x Time = 120 km/h x (25/60) h = 50 km
Now, let's calculate the displacement of the second plane from 10:05 am to 10:25 am.
Time elapsed: 10:25 am - 10:05 am = 20 minutes = 20/60 hours
Distance traveled by the second plane: Speed x Time = 90 km/h x (20/60) h = 30 km
Now, we have the following information:
- The first plane has traveled 50 km.
- The second plane has traveled 30 km.
To find the distance between the planes, we can use the Pythagorean theorem.
Distance = √((50^2) + (30^2))
Distance ≈ √(2500 + 900)
Distance ≈ √3400
Distance ≈ 58.31 km
Therefore, the planes are approximately 58.31 km apart at 10:25 am.
To find the distance between the two planes at 10.25 am, we need to determine the positions of both planes at that time.
Let's start by calculating the distance each plane has traveled from the airfield.
The first plane took off at 10.00 am and travels at a speed of 120 km/h on a bearing of N35W. Since it has been flying for 25 minutes (from 10.00 am to 10.25 am), it has covered a distance of:
Distance = Speed × Time
= 120 km/h × (25/60) h
= 50 km
The second plane took off at 10.05 am and flies at a speed of 90 km/h on a bearing of S80E. Since it has been flying for 20 minutes (from 10.05 am to 10.25 am), it has covered a distance of:
Distance = Speed × Time
= 90 km/h × (20/60) h
= 30 km
Now, we can find the position of each plane at 10.25 am relative to the airfield.
For the first plane, since it started flying towards N35W, and we know it traveled 50 km, we can construct a right triangle with the airfield at one corner, the bearing N35W, and the distance traveled as the hypotenuse.
To find the positions at 10.25 am, we can use trigonometry to find the northward and westward distances covered by the first plane.
Northward distance = Distance × cos(35°)
= 50 km × cos(35°)
= 41.13 km (approx.)
Westward distance = Distance × sin(35°)
= 50 km × sin(35°)
= 28.38 km (approx.)
For the second plane, since it started flying towards S80E, and we know it traveled 30 km, we can construct a right triangle with the airfield at one corner, the bearing S80E, and the distance traveled as the hypotenuse.
To find the positions at 10.25 am, we can use trigonometry to find the southward and eastward distances covered by the second plane.
Southward distance = Distance × cos(80°)
= 30 km × cos(80°)
= 6.47 km (approx.)
Eastward distance = Distance × sin(80°)
= 30 km × sin(80°)
= 29.92 km (approx.)
Now, we can find the horizontal and vertical distances between the two planes using the positions calculated above.
Horizontal distance = Westward distance + Eastward distance
= 28.38 km + 29.92 km
= 58.3 km (approx.)
Vertical distance = Northward distance + Southward distance
= 41.13 km + 6.47 km
= 47.6 km (approx.)
Finally, we can calculate the distance between the two planes at 10.25 am using the Pythagorean theorem.
Distance between the planes = √(Horizontal distance^2 + Vertical distance^2)
= √(58.3 km^2 + 47.6 km^2)
= √(3408.89 km^2 + 2267.36 km^2)
= √(5676.25 km^2)
= 75.4 km (approx.)
Therefore, the two planes are approximately 75.4 km apart at 10.25 am.