Answered

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A miniature quadcopter is located at xi = 1.50 m and yi = 4.70 m at t = 0 and moves with an average velocity having components vav, x = 2.70 m/s and vav, y = −2.50 m/s. What are the x-coordinate and y-coordinate (in m) of the quadcopter's position at t = 3.10 s?

Sagot :

Answer:

xf = 9.87 m and yf = -3.05 m

Explanation:

The x and y coordinated can be calculated using the following equation

xf = xi + Vav,x(t)

yf = yi + Vav,y(t)

Where xi, yi are the initial positions, Vav,x and Vav,y are the averages velocities and t is the time.

Replacing the given values, we get that the equations are

xf = 1.50 + 2.70t

yf = 4.70 - 2.50t

Now, we can calculate the position at t = 3.10s, replacing t by 3.10, so

xf = 1.50 + 2.70(3.10)

xf = 1.50 + 8.37

xf = 9.87 m

yf = 4.70 - 2.50(3.10)

yf = 4.70 - 7.75

yf = -3.05 m

Therefore, the x-coordinate and y-coordinate are xf = 9.87 m and yf = -3.05 m

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