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kondaur [170]
3 years ago
10

A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co

urse to 210° and continues in this direction for 1 hour. How far is the plane from the airport at the end of this time?
Mathematics
1 answer:
butalik [34]3 years ago
8 0

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

\displaystyle \vec{v}=\frac{\vec{x}}{t}

Where \vec x is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

\displaystyle \vec{x}=\vec{v}.t

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=\ km/h

The displacement of the plane in 2 hours is

\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)

\displaystyle \vec{x_1}=km

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=km/h

The displacement in 1 hour is

\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)

\displaystyle \vec{x_2}=km

The total displacement is the vector sum of both

\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+

\displaystyle \vec{x_t}=km

\displaystyle \vec{x_t}=

The distance from the airport is the module of the displacement:

\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}

\displaystyle |\vec{x_t}|=894.43\ km

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