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Chapter 3: Problem 38

A hawk is flying north at \(2.0 \mathrm{m} / \mathrm{s}\) with respect to the ground; 10.0 s later, it is flying south at \(5.0 \mathrm{m} / \mathrm{s} .\) What is its average acceleration during this time interval?

Short Answer

Expert verified
Answer: The average acceleration of the hawk is -0.7 m/s² (south direction).

Step by step solution

01

Note down the given data

The initial velocity (u) of the hawk is +2.0 m/s (north direction is positive). The final velocity (v) is -5.0 m/s. (south direction is negative). The time interval (t) is 10.0 s.
02

Define the formula

The average acceleration (a) is the change in velocity divided by the time interval taken. The formula for average acceleration (a) is given by: a = (v - u) / t
03

Substitute the given values

Now, substitute the given values in the formula. a = (-5.0 m/s - 2.0 m/s) / (10.0 s)
04

Calculate the average acceleration

Perform the calculation to find the average acceleration: a = (-7.0 m/s) / (10.0 s) a = -0.7 m/s² The hawk's average acceleration during this time interval is -0.7 m/s² (south direction).

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Most popular questions from this chapter

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