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.

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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