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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

The position of a particle moving along the \(x\) -axis varies with time according to the expression \(x=4 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. Evaluate the particle's position a) at \(t=2.00 \mathrm{~s}\). b) at \(2.00 \mathrm{~s}+\Delta t\) c) Evaluate the limit of \(\Delta x / \Delta t\) as \(\Delta t\) approaches zero, to find the velocity at \(t=2.00 \mathrm{~s}\).

A double speed trap is set up on a freeway. One police cruiser is hidden behind a billboard, and another is some distance away under a bridge. As a sedan passes by the first cruiser, its speed is measured to be \(105.9 \mathrm{mph}\). Since the driver has a radar detector, he is alerted to the fact that his speed has been measured, and he tries to slow his car down gradually without stepping on the brakes and alerting the police that he knew he was going too fast. Just taking the foot off the gas leads to a constant deceleration. Exactly 7.05 s later the sedan passes the second police cruiser. Now its speed is measured to be only \(67.1 \mathrm{mph}\), just below the local freeway speed limit. a) What is the value of the deceleration? b) How far apart are the two cruisers?

A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(3.00 \mathrm{~s}\) if its acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due west. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

The rate of continental drift is on the order of \(10.0 \mathrm{~mm} / \mathrm{yr}\). Approximately how long did it take North America and Europe to reach their current separation of about \(3000 \mathrm{mi}\) ?

A stone is thrown downward with an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\). The acceleration of the stone is constant and has the value of the free-fall acceleration, \(9.81 \mathrm{~m} / \mathrm{s}^{2} .\) What is the velocity of the stone after \(0.500 \mathrm{~s} ?\)
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