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Chapter 2: Problem 17

A car's velocity as a function of time is given by \(v_{x}(t)=\alpha+\beta t^{2},\) where \(\alpha=3.00 \mathrm{m} / \mathrm{s}\) and \(\beta=0.100 \mathrm{m} / \mathrm{s}^{3} .\) (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00\) s. (b) Calculate the instantaneous acceleration for \(t=0\) and \(t=5.00\) s. (c) Draw \(v_{x^{-}} t\) and \(a_{x^{-}}\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{s}.\)

Short Answer

Expert verified
(a) Average acceleration is \(0.50 \, \mathrm{m/s^2}\). (b) Instantaneous accelerations are \(0.00\) and \(1.00 \, \mathrm{m/s^2}\). (c) Graphs are quadratic for velocity and linear for acceleration.

Step by step solution

01

Calculate average acceleration

The average acceleration is defined as the change in velocity over a given time interval. The velocity function is given as \( v_{x}(t) = \alpha + \beta t^2 \). First, calculate \( v_{x}(0) = \alpha + \beta \cdot 0^2 = 3.00 \, \mathrm{m/s} \). Then, calculate \( v_{x}(5) = \alpha + \beta \cdot 5^2 = 3.00 + 0.100 \cdot 25 = 5.50 \, \mathrm{m/s} \). The change in velocity is \( \Delta v = v_{x}(5) - v_{x}(0) = 5.50 - 3.00 = 2.50 \, \mathrm{m/s} \). The time interval \( \Delta t = 5.00 - 0 = 5.00 \, \mathrm{s} \). Therefore, the average acceleration \( a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{2.50}{5.00} = 0.50 \, \mathrm{m/s^2} \).
02

Calculate instantaneous acceleration at t=0

Instantaneous acceleration is the derivative of the velocity function with respect to time. Differentiate \( v_{x}(t) = \alpha + \beta t^2 \) to get \( a_{x}(t) = \frac{d}{dt}(\alpha + \beta t^2) = 2\beta t \). Substitute \( t = 0 \) to get \( a_{x}(0) = 2\beta \cdot 0 = 0.00 \, \mathrm{m/s^2} \).
03

Calculate instantaneous acceleration at t=5.00s

Using the same derivative function \( a_{x}(t) = 2\beta t \), substitute \( t = 5.00 \) to get \( a_{x}(5.00) = 2 \cdot 0.100 \cdot 5.00 = 1.00 \, \mathrm{m/s^2} \).
04

Draw velocity vs. time graph

Plot a graph with time \( t \) on the x-axis and velocity \( v_{x}(t) \) on the y-axis. The equation for velocity is \( v_{x}(t) = 3.00 + 0.100 t^2 \). This is a quadratic function with the parabola opening upwards starting at \( 3.00 \, \mathrm{m/s} \). Calculate and plot points for \( t = 0, 1, 2, 3, 4, 5 \).
05

Draw acceleration vs. time graph

Plot a graph with time \( t \) on the x-axis and acceleration \( a_{x}(t) \) on the y-axis. The equation for acceleration is \( a_{x}(t) = 2\beta t = 0.200 t \). This function is a straight line starting at the origin and increasing linearly with time. Calculate and plot points for \( t = 0, 1, 2, 3, 4, 5 \).

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Key Concepts

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

Average Acceleration
In kinematics, average acceleration is the change in velocity divided by the time over which the change occurs. When dealing with this concept, imagine measuring how quickly a car speeds up from one moment to another.
For example, if a car's velocity goes from 3.00 m/s to 5.50 m/s over 5 seconds, its average acceleration can be calculated as:
  • Change in velocity: 5.50 m/s - 3.00 m/s = 2.50 m/s
  • Time interval: 5 seconds
  • Average acceleration: \( a_{\text{avg}} = \frac{2.50 \text{ m/s}}{5 \text{ s}} = 0.50 \text{ m/s}^2 \)
This tells us that on average, the velocity increases by 0.50 m/s each second over the period considered.
Instantaneous Acceleration
Instantaneous acceleration is like pressing the fast-forward button on average acceleration. Instead of considering the overall change over a time span, it looks at how velocity is changing at a specific moment.
To find this, you differentiate the velocity function with respect to time. For a car with velocity function \( v_{x}(t) = \alpha + \beta t^2 \):
  • The derivative \( a_{x}(t) = 2\beta t \)
At \( t=0 \):
  • \( a_{x}(0) = 2\beta \cdot 0 = 0.00 \text{ m/s}^2 \)
At \( t=5.00 \text{ s} \):
  • \( a_{x}(5.00) = 2\cdot 0.100 \cdot 5.00 = 1.00 \text{ m/s}^2 \)
This tells us how swiftly the car's speed is changing precisely at those specific times.
Velocity-Time Graph
A velocity-time graph provides a visual representation of how velocity changes over time. It helps us better understand motion patterns such as speeding up, slowing down, or maintaining velocity.
For the function \( v_{x}(t) = 3.00 + 0.100 t^2 \), this graph would show a parabola opening upwards. The starting point is 3.00 m/s, which is the car's initial velocity. As time progresses, the velocity increases in a nonlinear fashion due to the squared term \( t^2 \).
Key points that would be plotted include:
  • \( v_{x}(0) = 3.00 \text{ m/s} \)
  • \( v_{x}(1) = 3.10 \text{ m/s} \)
  • \( v_{x}(2) = 3.40 \text{ m/s} \)
  • and up to \( v_{x}(5) = 5.50 \text{ m/s} \)
This type of graph is crucial for visualizing how an object's speed changes over a given period.
Acceleration-Time Graph
The acceleration-time graph illustrates how acceleration varies over time. It shows whether an object speeds up or slows down and how rapidly it happens.
For a given acceleration function \( a_{x}(t) = 2\beta t \), the graph is a straight line starting at the origin and rising linearly. This indicates constant change in acceleration鈥攁 major clue into the dynamics of motion.
For our specific example:
  • At \( t=0 \), acceleration is \( 0 \text{ m/s}^2 \)
  • At \( t=1 \), acceleration is \( 0.200 \text{ m/s}^2 \)
  • By \( t=5 \), acceleration is \( 1.00 \text{ m/s}^2 \)
The straight line suggests that however the speed changes, it's a consistent increase over time, directly related to how fast the time goes by.

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Most popular questions from this chapter

Trip Home. You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 \(\mathrm{km} / \mathrm{h}\) (65 \(\mathrm{mi} / \mathrm{h} ),\) and the trip takes 2 \(\mathrm{h}\) and 20 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 \(\mathrm{km} / \mathrm{h}(43 \mathrm{mi} / \mathrm{h})\) How much longer does the trip take?

The Fastest (and Most Expensive) Car! The table shows test data for the Bugatti Veyron, the fastest car made. The car is moving in a straight line (the \(x\)-axis). \(\begin{array}{llll}{\text { Time }(s)} & {0} & {2.1} & {20.0} & {53} \\\ {\text { Speed (mijh) }} & {0} & {60} & {200} & {253}\end{array}\) (a) Make a \(v_{x}-t\) graph of this car's velocity \((\) in \(\mathrm{mi} / \mathrm{h})\) as a function of time. Is its acceleration constant? (b) Calculate the car's average acceleration ( in \(\mathrm{m} / \mathrm{s}^{2} )\) between ( i 0 and \(2.1 \mathrm{s} ;\) (ii) 2.1 \(\mathrm{s}\) and 20.0 \(\mathrm{s}\); (iii) 20.0 s and 53 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300 will be built, it runs out of gas in 12 minutes at top speed, and it costs \(\$ 1.25\) million!)

Dan gets on Interstate Highway \(\mathrm{I}-80\) at Seward, Nebraska, and drives due west in a straight line and at an average velocity of magnitude 88 \(\mathrm{km} / \mathrm{h}\) . After traveling 76 \(\mathrm{km}\) , he reaches the Aurora exit (Fig. \(\mathrm{P} 2.63 ) .\) Realizing he has gone too far, he turns around and drives due east 34 \(\mathrm{km}\) back to the York exit at an average velocity of magnitude 72 \(\mathrm{km} / \mathrm{h}\) . For his whole trip from Seward to the York exit, what are (a) his average speed and (b) the magnitude of his average velocity?

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s} .\) If you start 0.90 m behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

(a) If a flea can jump straight up to a height of \(0.440 \mathrm{m},\) what is its initial speed as it leaves the ground? (b) How long is it in the air?
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