Help with Homework Problem on Acceleration

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SUMMARY

The Australian emu holds the land speed record of 13.9 m/s (31 mi/h) and decelerates from this speed to 10.5 m/s over 2.7 seconds. The direction of the bird's acceleration is due north, opposite to its initial southward velocity. Assuming constant acceleration, the bird's velocity after an additional 3.9 seconds is calculated to be 6.6 m/s due north. The acceleration is determined to be -0.4 m/s², indicating a decrease in speed.

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shawonna23
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The land speed record of 13.9 m/s (31 mi/h) for birds is held by the Australian emu. An emu running due south in a straight line at this speed slows down to a speed of 10.5 m/s in 2.7 s.

What is the direction of the bird's acceleration?

Assuming that the acceleration remains the same, what is the bird's velocity after an additional 3.9 s has elapsed?
 
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shawonna23 said:
The land speed record of 13.9 m/s (31 mi/h) for birds is held by the Australian emu. An emu running due south in a straight line at this speed slows down to a speed of 10.5 m/s in 2.7 s.

What is the direction of the bird's acceleration?

Assuming that the acceleration remains the same, what is the bird's velocity after an additional 3.9 s has elapsed?

direction of bird's acceleration: Well obviously it's not due south, otherwise it would be accelerating. Hmm i wonder...

10.5 = 13.9 + a(2.7)

a = - (3.4/2.7)m/s^2<------------ what do you think the negative sign signifies?

for the second part. Just use vf = vi + at
 


The direction of the bird's acceleration would be in the opposite direction of its velocity, which in this case is due south. This means that the acceleration is in the north direction.

If the acceleration remains the same, the bird's velocity after an additional 3.9 s would be 6.6 m/s due north. This can be calculated by subtracting the acceleration of 0.4 m/s^2 (calculated by dividing the change in velocity, 2.7 m/s, by the change in time, 2.7 s) from the initial velocity of 10.5 m/s and then multiplying it by the additional time of 3.9 s. This would result in a velocity of 6.6 m/s due north.
 

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