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An object has an acceleration of \(+1.2 \mathrm{cm} / \mathrm{s}^{2} .\) At \(t=4.0 \mathrm{s},\) its velocity is \(-3.4 \mathrm{cm} / \mathrm{s} .\) Determine the object's velocities at \(t=1.0 \mathrm{s}\) and \(t=6.0 \mathrm{s}\).

Short Answer

Expert verified
At t = 1.0 s, the object's velocity is \(v = -7.0 \mathrm{cm}/\mathrm{s}\), and at t = 6.0 s, its velocity is \(v = -1.0 \mathrm{cm}/\mathrm{s}\).

Step by step solution

01

Calculate the time intervals

First, we need to find the time intervals for t = 1.0 s and t = 6.0 s with respect to t = 4.0 s. This is because the given velocity (-3.4 cm/s) is at t = 4.0 s. For t = 1.0 s: Time interval = 1.0 s - 4.0 s = -3.0 s For t = 6.0 s: Time interval = 6.0 s - 4.0 s = 2.0 s
02

Use the kinematics equation to find the velocities

Now we will use the kinematics equation, v = v0 + at, to find the velocities at t = 1.0 s and t = 6.0 s. Given: - Initial velocity (v0) = -3.4 cm/s (at t = 4.0 s) - Acceleration (a) = 1.2 cm/s² For t = 1.0 s (time interval = -3.0 s): v = -3.4 cm/s + (1.2 cm/s²)(-3.0 s) v = -3.4 cm/s - 3.6 cm/s v = -7.0 cm/s For t = 6.0 s (time interval = 2.0 s): v = -3.4 cm/s + (1.2 cm/s²)(2.0 s) v = -3.4 cm/s + 2.4 cm/s v = -1.0 cm/s So, at t = 1.0 s, the object's velocity is -7.0 cm/s, and at t = 6.0 s, its velocity is -1.0 cm/s.

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Key Concepts

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

Acceleration
Acceleration is a fundamental concept in kinematics, which describes how an object's velocity changes over time. It measures the rate of change of velocity per unit of time. In this exercise, we have an acceleration of \(+1.2 \, \mathrm{cm/s^2}\), which means every second, the velocity of the object increases by \(+1.2 \, \mathrm{cm/s}\).
Understanding acceleration helps predict how fast or slow an object's speed is changing. When the acceleration is positive, it means the object is speeding up in the direction of motion. Conversely, a negative acceleration indicates the object is slowing down.
For example:
  • If the velocity of an object is negative, like \(-3.4 \, \mathrm{cm/s}\), a positive acceleration decreases the negativity (makes it less negative or more positive over time).
  • Over time, this change can be calculated using the kinematics equations, allowing us to know the velocity at different moments.
Velocity Calculation
Velocity calculation involves determining the current speed and direction of an object in motion. It is derived from knowing the object's initial velocity and how that velocity changes over time due to acceleration.
In our exercise, we start with an initial velocity of \(-3.4 \, \mathrm{cm/s}\) at \(t = 4.0 \, \mathrm{s}\). To calculate velocity at different times like \(t = 1.0 \, \mathrm{s}\) and \(t = 6.0 \, \mathrm{s}\), we must first determine how far these are from our initial time.To do this:
  • Calculate time intervals: For \(t = 1.0 \, \mathrm{s}\), the interval is \(-3.0 \, \mathrm{s}\) and for \(t = 6.0 \, \mathrm{s}\), the interval is \(+2.0 \, \mathrm{s}\).
  • Apply the kinematics equation by plugging these intervals into the formula \(v = v_0 + at\) to find the new velocities.
By understanding these intervals and how they relate to acceleration, we determine the velocities: \(-7.0 \, \mathrm{cm/s}\) at \(t = 1.0 \, \mathrm{s}\) and \(-1.0 \, \mathrm{cm/s}\) at \(t = 6.0 \, \mathrm{s}\). This shows how the object changes speed and direction over time.
Kinematics Equation
The kinematics equation is a simple yet powerful tool for predicting an object's future velocity. It is based on the relationship between initial velocity, constant acceleration, and time. The formula used is \(v = v_0 + at\), where:
  • \(v\) is the final velocity you're trying to find.
  • \(v_0\) represents the initial velocity of the object.
  • \(a\) is the acceleration, which remains constant here.
  • \(t\) is the time elapsed since the initial observation.
Given the provided initial conditions—velocity at \(t = 4.0 \, \mathrm{s}\) and the object's acceleration—you can find how the velocity changes at different times by plugging numbers into this equation.
This method allows us to accurately predict velocities, like \(-7.0 \, \mathrm{cm/s}\) and \(-1.0 \, \mathrm{cm/s}\) for the object at \(t = 1.0 \, \mathrm{s}\) and \(t = 6.0 \, \mathrm{s}\) respectively. This equation is especially useful because it requires only straightforward arithmetic to solve complex motion problems, making it accessible for students learning kinematics.

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Most popular questions from this chapter

The velocity of a particle moving along the \(x\) -axis varies with time according to \(v(t)=A+B t^{-1},\) where \(A=\) \(2 \mathrm{m} / \mathrm{s}, B=0.25 \mathrm{m},\) and \(1.0 \mathrm{s} \leq t \leq 8.0 \mathrm{s} .\) Determine the acceleration and position of the particle at \(t=2.0 \mathrm{s}\) and \(t=\) 5.0 s. Assume that \(x(t=1 \mathrm{s})=0\).

Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December \(10,1954,\) Stapp rode a rocket sled, accelerating from rest to a top speed of \(282 \mathrm{m} / \mathrm{s}\) (1015 \(\mathrm{km} / \mathrm{h}\) ) in \(5.00 \mathrm{s}\) and was brought jarringly back to rest in only 1.40 s. Calculate his (a) acceleration in his direction of motion and (b) acceleration opposite to his direction of motion. Express each in multiples of \(g\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)\) by taking its ratio to the acceleration of gravity.

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.

An ambulance driver is rushing a patient to the hospital. While traveling at \(72 \mathrm{km} / \mathrm{h}\), she notices the traffic light at the upcoming intersections has turned amber. To reach the intersection before the light turns red, she must travel \(50 \mathrm{m}\) in \(2.0 \mathrm{s}\). (a) What minimum acceleration must the ambulance have to reach the intersection before the light turns red? (b) What is the speed of the ambulance when it reaches the intersection?

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of \(8.00 \mathrm{m} / \mathrm{s}\). (b) How long a time would it take to reach the ground if it is thrown straight down with the same speed?

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