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Chapter 2: Problem 30

The position of a particle moving along the \(x\) -axis is given by \(x=\left(11+14 t-2.0 t^{2}\right),\) where \(t\) is in seconds and \(x\) is in meters. What is the average velocity during the time interval from \(t=1.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s} ?\)

Short Answer

Expert verified
Answer: The average velocity during the time interval from \(t=1.0\text{ s}\) to \(t=4.0\text{ s}\) is \(4.0 \frac{\text{m}}{\text{s}}\).

Step by step solution

01

Calculate Position at the Starting and Ending of the Time Interval

Using the position function, find the position of the particle at \(t=1.0s\) and \(t=4.0s\): \(x(1.0) = 11 + 14(1.0) - 2.0(1.0)^2\) \(x(4.0) = 11 + 14(4.0) - 2.0(4.0)^2\)
02

Compute Positions

Calculate the positions at the given times: \(x(1.0) = 11 + 14 - 2 = 23\text{ m}\) \(x(4.0) = 11 + 56 - 32 = 35\text{ m}\)
03

Find Displacement

Subtract the initial position from the final position to find the displacement: \(\Delta x = x(4.0) - x(1.0) = 35 - 23 = 12\text{ m}\)
04

Calculate Time Interval

Find the time interval between the given times: \(\Delta t = 4 - 1 = 3\text{ s}\)
05

Compute Average Velocity

Divide the displacement by the time interval to find the average velocity: \(v_\text{avg} = \frac{\Delta x}{\Delta t} = \frac{12}{3} = 4.0 \frac{\text{m}}{\text{s}}\) The average velocity during the time interval from \(t=1.0\text{ s}\) to \(t=4.0\text{ s}\) is \(4.0 \frac{\text{m}}{\text{s}}\).

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Most popular questions from this chapter

A car traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) applies the brakes and decelerates uniformly at a rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) a) How far does it travel in \(3.0 \mathrm{~s}\) ? b) What is its velocity at the end of this time interval? c) How long does it take for the car to come to a stop? d) What distance does the car travel before coming to a stop?

A ball is thrown directly downward, with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\), from a height of \(50.0 \mathrm{~m}\). After what time interval does the ball strike the ground?

You and a friend are standing at the edge of a snowcovered cliff. At the same time, you throw a snowball straight upward with a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) over the edge of the cliff and your friend throws a snowball straight downward over the edge of the cliff with the same speed. Your snowball is twice as heavy as your friend's. Neglecting air resistance, which snowball will hit the ground first, and which will have the greater speed?

A car travels at 22.0 mph for 15.0 min and 35.0 mph for \(30.0 \mathrm{~min}\). How far does it travel overall? a) \(23.0 \mathrm{~m}\) b) \(3.70 \cdot 10^{4} \mathrm{~m}\) c) \(1.38 \cdot 10^{3} \mathrm{~m}\) d) \(3.30 \cdot 10^{2} \mathrm{~m}\)

Two athletes jump straight up. Upon leaving the ground, Adam has half the initial speed of Bob. Compared to Adam, Bob jumps a) 0.50 times as high. b) 1.41 times as high. c) twice as high. d) three times as high. e) four times as high.
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