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Elena L [17]
2 years ago
15

A plane lands on a runway with a speed of 115 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2

) and direction of the plane's average acceleration during this time interval
Physics
1 answer:
marin [14]2 years ago
4 0

Answer:

<em> The planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I Unit of acceleration is m/s². Acceleration is a vector quantity because it can be represented both in magnitude and in direction.

Acceleration can be represented mathematically as

a = v/t.................................... Equation 1

Where a = acceleration, v = velocity, t= time.

<em>Given: v = 115 m/s, t = 13.0 s</em>

<em>Substituting these values into equation 1</em>

<em>a = 115/13</em>

<em>a = 8.846 m/s² moving east</em>

<em>Thus the planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

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Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

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\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

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