The speed of the two planes is 52.7 m/s, with the first plane travelling at this...

The speed of the two planes is 52.7 m/s, with the first plane travelling at this velocity in the y-direction and the second plane travelling with this speed in the x-direction. The initial separation of the planes is 30.5 m. Find the time at which the separation of the planes is at its minimum.

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Expert Solution

Its given that the initial distance between both the planes is 30.5m ann speed of both the planes is 52.7m/s. Let the plane A be at the coordinates (0,0) initially and its moving along positive x axis and the coordinates of plane B (0,-30.5) and moving along positive y axis.

Let us try to solve this with the concept of relative velocity.

Assuming plane A is stationary, relative velocity of plane B with respect to plane A is 52.7(-i+j)m/s from its initial position (0,-30.5).

So plane A is stationary and plane B started moving with speed 52.7(-i+j)m/s in third quadrant.

We can clearly see that the distance between them will be minimum when coordinates of B will be (-15.25,-15.25).

Since plane B is moving with velocity of 52.7m/s in both negative x axis and positive y axis and initially it was at -30.5m on y axis, So the time required to get to -15.25 on y axis with velocity 52.7m/s along y axis is

15.25/52.7 = 0.289sec

So the answer is 0.289sec

genius_generous answered 1 year ago

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