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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

A particle has a constant acceleration of \(5.0 \mathrm{m} / \mathrm{s}^{2}\) to the east. At time \(t=0,\) it is \(2.0 \mathrm{m}\) east of the origin and its velocity is \(20 \mathrm{m} / \mathrm{s}\) north. What are the components of its position vector at \(t=2.0 \mathrm{s} ?\)
The range \(R\) of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at \(t=0\) with initial speed \(v_{i}\) at an angle \(\theta\) above the horizontal. (a) Find the time \(t\) at which the projectile returns to its original altitude. (b) Show that the range is \(R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}\) [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of \(\theta\) gives the maximum range? What is this maximum range?
A car is driving directly north on the freeway at a speed of $110 \mathrm{km} / \mathrm{h}$ and a truck is leaving the freeway driving \(85 \mathrm{km} / \mathrm{h}\) in a direction that is \(35^{\circ}\) west of north. What is the velocity of the truck relative to the car?
Show that for a projectile launched at an angle of \(45^{\circ}\) the maximum height of the projectile is one quarter of the range (the distance traveled on flat ground).
John drives \(16 \mathrm{km}\) directly west from Orion to Chester at a speed of \(90 \mathrm{km} / \mathrm{h},\) then directly south for \(8.0 \mathrm{km}\) to Seiling at a speed of \(80 \mathrm{km} / \mathrm{h}\), then finally $34 \mathrm{km}\( southeast to Oakwood at a speed of \)100 \mathrm{km} / \mathrm{h}$. Assume he travels at constant velocity during each of the three segments. (a) What was the change in velocity during this trip? [Hint: Do not assume he starts from rest and stops at the end.] (b) What was the average acceleration during this trip?
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