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(a) If a particle's position is given by \(x=4-12 t+3 t^{2}\) (where \(t\) is in seconds and \(x\) is in meters), what is its velocity at \(t=1 \mathrm{~s} ?(\mathrm{~b})\) Is it moving in the positive or negative direction of \(x\) just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time \(t ;\) if not, answer no. (f) Is there a time after \(t=3 \mathrm{~s}\) when the particle is moving in the negative direction of \(x ?\) If so, give the time \(t ;\) if not, answer no.

Short Answer

Expert verified
At t=1s, velocity is -6 m/s, speed is 6 m/s, and speed is decreasing. Velocity is zero at t=2s. After t=3s, velocity is not negative.

Step by step solution

01

Find the Velocity Function

To find the velocity function, differentiate the given position function. The position function is given as \[ x(t) = 4 - 12t + 3t^2. \]The velocity is the derivative of \( x(t) \) with respect to \( t \):\[ v(t) = \frac{dx}{dt} = -12 + 6t. \]
02

Calculate Velocity at t=1s

Substitute \( t=1 \) into the velocity function:\[ v(1) = -12 + 6 \times 1 = -6 \, \text{m/s}. \]
03

Determine Direction of Motion

The sign of the velocity at \( t=1 \) is negative, which means the particle is moving in the negative direction of \( x \) at this instant.
04

Calculate Speed at t=1s

Speed is the absolute value of velocity. Thus,\[ \text{Speed} = |-6| = 6 \, \text{m/s}. \]
05

Determine if Speed is Increasing or Decreasing

Since the velocity is negative and the derivative of velocity \( (6t - 12) \) indicates an increasing velocity for \( t>2 \), the speed is decreasing at \( t=1 \), as velocity's magnitude is still decreasing towards zero.
06

Check for Zero Velocity

Set the velocity function \( v(t) = 0 \) to check for zero velocity:\[ -12 + 6t = 0 \Rightarrow t = 2 \, \text{seconds}. \]
07

Check for Negative Direction After t=3s

Evaluate \( v(t) = -12 + 6t \) after \( t=3 \):For \( t>3 \), \( v(t) = -12 + 6t > 0 \), implying no time after \( t=3s \) when the velocity is negative. Thus, answer: no.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity
Velocity is a key concept in kinematics, representing how fast an object changes its position over time. It is a vector quantity, which means it has both magnitude and direction. Velocity tells you not just how fast something is moving, but also in which direction. This is important because two objects could move at the same speed yet have completely different velocities if they are moving in opposite directions.

To find the velocity of a particle from its position function, you need to differentiate the position function with respect to time. For example, given a position function like \( x(t) = 4 - 12t + 3t^2 \), the velocity \( v(t) \) is found by taking the derivative:
  • \( v(t) = \frac{dx}{dt} = -12 + 6t \)
This equation \( v(t) = -12 + 6t \) tells us the velocity of the particle at any time \( t \). By substituting different values of \( t \), one can easily find out how the particle's velocity changes over time.
Position function
The position function describes how an object's location changes over time. It provides a direct relationship between time and the particle's position. This function is crucial as it helps determine other important kinematic quantities like velocity and acceleration.

In our exercise, the position function is given as \( x(t) = 4 - 12t + 3t^2 \), where \( x \) represents the particle's position in meters and \( t \) is time in seconds. With this quadratic equation, we can analyze how the position of the particle varies as time progresses. If you plot the position function, it will typically manifest as a parabolic curve, indicating how the object moves through space.

Understanding the position function is essential because it is from this function that we derive the velocity function, as well as subsequent calculations of acceleration and more.
Speed
Speed, often confused with velocity, is a scalar quantity. Unlike velocity, speed does not have a direction—it only provides information on how fast an object is moving. This means speed is essentially the magnitude of velocity.

In the context of our particle, if the velocity at a certain point is given as \( v = -6 \, \text{m/s} \), the speed at that point would simply be the absolute value of velocity:
  • Speed = \( |-6| = 6 \, \text{m/s} \)
Speed is always a non-negative value. It is useful when you need to understand the pure moving capability of an object without needing to account for direction. Even if the velocity changes direction, the speed can remain constant if the magnitude doesn’t vary.
Instantaneous velocity
Instantaneous velocity refers to the velocity of an object at a specific moment in time. It is an important concept in kinematics because it gives precise detail about the motion of an object at any particular instant.

To determine the instantaneous velocity, you evaluate the velocity function at a specific time. For example, in our case, to find the velocity at \( t = 1 \, \text{second} \), you would calculate:
  • \( v(1) = -12 + 6 \times 1 = -6 \, \text{m/s} \)
The result \( -6 \, \text{m/s} \) indicates the particle's velocity at exactly one second, giving both the speed and direction of movement. Instantaneous velocity highlights the dynamic nature of movement, which can vary from moment to moment, unlike average velocity, which overviews a broader time period.

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Most popular questions from this chapter

A bolt is dropped from a bridge under construction, falling \(90 \mathrm{~m}\) to the valley below the bridge. (a) In how much time does it pass through the last \(20 \%\) of its fall? What is its speed (b) when it begins that last \(20 \%\) of its fall and \((\mathrm{c})\) when it reaches the valley beneath the bridge?

You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of \(v_{0}=55 \mathrm{~km} / \mathrm{h} ;\) your best deceleration rate has the magnitude \(a=5.18 \mathrm{~m} / \mathrm{s}^{2}\). Your best reaction time to begin braking is \(T=0.75 \mathrm{~s}\). To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at \(55 \mathrm{~km} / \mathrm{h}\) if the distance to the intersection and the duration of the yellow light are (a) \(40 \mathrm{~m}\) and \(2.8 \mathrm{~s}\), and (b) \(32 \mathrm{~m}\) and \(1.8 \mathrm{~s}\) ? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).

In an arcade video game, a spot is programmed to move across the screen according to \(x=9.00 t-0.750 t^{3}\), where \(x\) is distance in centimeters measured from the left edge of the screen and \(t\) is time in seconds. When the spot reaches a screen edge, at either \(x=0\) or \(x=15.0 \mathrm{~cm}, t\) is reset to 0 and the spot starts moving again according to \(x(t) .\) (a) At what time after starting is the spot instantaneously at rest? (b) At what value of \(x\) does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time \(t>0\) does it first reach an edge of the screen?

An object falls a distance \(h\) from rest. If it travels \(0.50 h\) in the last \(1.00 \mathrm{~s}\), find (a) the time and (b) the height of its fall. (c) Explain the physically unacceptable solution of the quadratic equation in \(t\) that you obtain.

When the legal speed limit for the New York Thruway was increased from \(55 \mathrm{mi} / \mathrm{h}\) to \(65 \mathrm{mi} / \mathrm{h}\), how much time was saved by a motorist who drove the \(700 \mathrm{~km}\) between the Buffalo entrance and the New York City exit at the legal speed limit?

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