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Chapter 2: Problem 43

A particle starts from rest at \(x=0\) and moves for \(20 \mathrm{~s}\) with an acceleration of \(+2.0 \mathrm{~cm} / \mathrm{s}^{2}\). For the next \(40 \mathrm{~s}\), the acceleration of the particle is \(-4.0 \mathrm{~cm} / \mathrm{s}^{2} .\) What is the position of the particle at the end of this motion?

Short Answer

Expert verified
Answer: The final position (displacement) of the particle at the end of its motion is -1200 cm.

Step by step solution

01

Find the final velocity at the end of the first stage

In this step, we will use the formula \(v = v_0 + at\) to find the final velocity at the end of the first stage, where \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. The initial velocity \(v_0\) is 0 because the particle starts from rest, the acceleration \(a\) is 2.0 cm/s², and the time \(t\) is 20s. \(v = 0 + (2.0 \mathrm{~cm/s^2})(20 \mathrm{~s}) = 40\mathrm{~cm/s}\) The final velocity at the end of the first stage is 40 cm/s.
02

Calculate the displacement at the end of the first stage

Now we will use the formula \(x = x_0 + v_0t + \frac{1}{2}at^2\) to find the displacement at the end of the first stage. The initial position \(x_0\) is 0, the initial velocity \(v_0\) is 0, the acceleration \(a\) is 2.0 cm/s², and the time \(t\) is 20s. \(x = 0 + 0 + \frac{1}{2}(2.0 \mathrm{~cm/s^2})(20 \mathrm{~s})^2 = 400\mathrm{~cm}\) The displacement at the end of the first stage is 400 cm.
03

Find the final velocity at the end of the second stage

At the start of the second 40s interval, the velocity of the particle is \(40\mathrm{~cm/s}\) which we calculated earlier. We can use the formula \(v = v_0 + at\) to find the final velocity at the end of the second stage, where \(v_0\) is the initial velocity (40 cm/s), \(a\) is the acceleration (-4.0 cm/s²), and \(t\) is the time (40s). \(v = 40\mathrm{~cm/s} + (-4.0\mathrm{~cm/s^2})(40\mathrm{~s}) = -120\mathrm{~cm/s}\) The final velocity at the end of the second stage is -120 cm/s.
04

Calculate the displacement at the end of the second stage

Now we will use the formula \(x = x_0 + v_0t + \frac{1}{2}at^2\) to find the displacement at the end of the second stage. The initial position \(x_0\) is 0, the initial velocity \(v_0\) is 40 cm/s, the acceleration \(a\) is -4.0 cm/s², and the time \(t\) is 40s. \(x = 0 + (40\mathrm{~cm/s})(40\mathrm{~s}) + \frac{1}{2}(-4.0 \mathrm{~cm/s^2})(40 \mathrm{~s})^2 = 1600 - 3200 = -1600\mathrm{~cm}\) The displacement at the end of the second stage is -1600 cm.
05

Calculate the total displacement

Finally, we will add the displacement of the first stage (400 cm) and the displacement of the second stage (-1600 cm) to find the total displacement. Total displacement = Displacement of the first stage + Displacement of the second stage Total displacement = 400 cm + (-1600 cm) = -1200 cm The final position of the particle at the end of this motion is -1200 cm.

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Most popular questions from this chapter

Runner 1 is standing still on a straight running track. Runner 2 passes him, running with a constant speed of \(5.1 \mathrm{~m} / \mathrm{s} .\) Just as runner 2 passes, runner 1 accelerates with a constant acceleration of \(0.89 \mathrm{~m} / \mathrm{s}^{2} .\) How far down the track does runner 1 catch up with runner \(2 ?\)

The velocity as a function of time for a car on an amusement park ride is given as \(v=A t^{2}+B t\) with constants \(A=2.0 \mathrm{~m} / \mathrm{s}^{3}\) and \(B=1.0 \mathrm{~m} / \mathrm{s}^{2} .\) If the car starts at the origin, what is its position at \(t=3.0\) s?

A car slows down from a speed of \(31.0 \mathrm{~m} / \mathrm{s}\) to a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) over a distance of \(380 . \mathrm{m} .\) a) How long does this take, assuming constant acceleration? b) What is the value of this acceleration?

A runner of mass 57.5 kg starts from rest and accelerates with a constant acceleration of \(1.25 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(6.3 \mathrm{~m} / \mathrm{s}\). She then continues running with this constant velocity. a) How far has she run after 59.7 s? b) What is the velocity of the runner at this point?

A train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2},\) and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?
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