A space ship is being piloted in deep space. The ships initial velocity is in th

Question

A space ship is being piloted in deep space. The ships initial velocity is in the x-direction 1.7km/s when it enters an asteroid field. Because a speeding asteroid is headed torwards the ship, the pilot needs to make an emergency maneuver. In 5.0 seconds the ship is turned and moving at a new velocity of -.5km/s in the x-direction and .5km/s in the y-direction

a) Draw a diagram depicting the beginning and end of the maneuver. Include the initial and final velocity vectors.

b) What is the average acceleration vector that the ship experienced in those 5 seconds?

c) If the magnitude of acceleration for a body at the surface of the earth due to gravity is g= 9.81 m/s^2 , how does this compare with the acceleration of the ships maneuver? Is this maneuver realistic? (Typically a human will pass out if accelerated at about 5g’s for an extended period of time)

d) Assuming a constant acceleration, what is the ships displacement from the start of the maneuver to the end of the maneuver?

Explanation / Answer

Here we have initial velocity in x direction, ux = 1.7 km/s and in y direction uy=0

Final veocity vx= -.5 and vy= .5 and t = 5 sec

Now from v = u + at, we have,

vx= ux+axt => ax=-2.2/5 = -.44km/s2 = -440 m/s2

vy= uy+ayt => ay= .1km/s2 = 100m/s2

Therefore, magnitude of average acceleration a = sqrt(ax2+ay2) = 451.22 m/s2

The acceleration is too high as compared to the acceleration due to gravity. Hence the maneuver is not realistic.

Now, displacement s = ut+.5at2 = (1700* 5)+(.5 * 451.22 * 25) = 14140.25 m = 14.14 km

Diagram would be like the following:

beginning of the maneuver, velocity vector was in the positive x axis i.e 1st quadrant

end of the maneuver, velocity vector was in the negative x axis and positive y axis i.e 2nd quadrant

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