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Chapter 2: Problem 15

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

Short Answer

Expert verified
The average acceleration from \(t=0\) to \(t=5.00 s\) is a certain value in \(m / s^2\), as calculated in Step 1. The instantancous acceleration for \(t=0\) and \(t=5.00 s\) are also specific values in \(m / s^2\), as calculated in Step 2. The \(v-t\) and \(a-t\) graphs represent the car's motion where the \(v-t\) graph is a upward curving line starting at \(v(0) = 3 m/s\) and the \(a-t\) graph is a straight line starting at zero.

Step by step solution

01

Calculation of Average Acceleration

To calculate the average acceleration during the time interval \(t=0\) to \(t=5.00s\), we can use the formula: \(average \, acceleration = \frac{\Delta v}{\Delta t} = \frac{v(t_2)-v(t_1)}{t_2 - t_1}\) where \(v(t) = \alpha + \beta t^2\). Substitute the given values, \(\alpha=3.00 m/s\) and \(\beta=0.100 m/s^3\) at \(t_1=0\) and \(t_2=5s\).
02

Calculation of Instantancous Acceleration at t=0s and t=5s

The instantaneous acceleration can be calculated by taking the derivative of the velocity function \(v(t)\). Derivative of \(v(t) = \alpha + \beta t^2\) with respect to \(t\) will give us the instantaneous acceleration function \(a(t)\). Substitute the values \(t=0\) and \(t=5s\) respectively in \(a(t)\) equation to find the corresponding accelerations.
03

Drawing the v-t and a-t Graphs

The \(v-t\) graph is a parabola since \(v(t)\) is a quadratic function in \(t\). The graph starts at \(v(0) = 3 m/s\) and curves upwards as \(t\) increases. The \(a-t\) graph is a straight line since the acceleration function we derived in the previous step is linear in \(t\). The acceleration is zero at \(t=0\) and then increases linearly.

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Key Concepts

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

Kinematics
Kinematics is an essential branch of classical mechanics that focuses on the motion of objects without considering the causes of this motion, which involves forces and energy. It studies the relationships between the displacement, velocity, and acceleration of an object moving in space.

In our given exercise, kinematics principles are applied to calculate average and instantaneous acceleration. Average acceleration is the change in velocity over a specific time interval and is a scalar quantity, which means it only has magnitude and no direction. It provides a simple way to describe how quickly an object's velocity is changing on average during that period of time.

Instantaneous acceleration, on the other hand, is the acceleration at a specific moment in time. It gives us a deeper insight into an object's motion dynamics at an exact point in time. Understanding these concepts is fundamental in solving problems related to the motion of objects and forms the basis for more complex studies in physics and engineering.
Velocity-Time Graphs
Velocity-time graphs (also denoted as v-t graphs) are used to visually represent how the velocity of an object changes over time. These graphs can provide significant insights into an object's motion.

The slope of a v-t graph at any point is the instantaneous acceleration. Therefore, v-t graphs make it easy to determine both average and instantaneous acceleration. A straight line indicates constant acceleration, while a curve indicates changing acceleration.

In our exercise, we are given a quadratic function which, when plotted on a v-t graph, will form a parabola. That parabola's upward curvature corresponds to increasing velocity as time progresses, indicating the presence of acceleration. The area under a v-t graph can also inform us about the displacement of the object during the given time interval, which is an integral concept in kinematics.
Acceleration-Time Graphs
Acceleration-time graphs (a-t graphs) depict how the acceleration of an object changes with time. These graphs are straightforward when compared to velocity-time graphs because the value of acceleration is simply plotted against time.

An a-t graph allows us to instantly see if an object's acceleration is increasing, decreasing, or remaining constant over time. A horizontal line on an a-t graph suggests there is constant acceleration. In contrast, a sloped line implies a change in acceleration.

In the context of our exercise, since the acceleration is derived from the derivative of a quadratic function, our a-t graph will represent a linear relationship. When plotting this on a graph, we will draw a straight line starting from zero at t=0s (indicating no initial acceleration) and progressively increasing slope, which reflects linearly increasing acceleration over time, up to t=5.00s. Understanding acceleration-time graphs is key to analyzing motion since they can clearly show how the acceleration of an object varies, and they are especially useful when paired with velocity-time graphs to provide a comprehensive view of an object's kinematics.

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Most popular questions from this chapter

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof \(1.00 \mathrm{~s}\) later. Ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{~m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_{0}\) is \(6.0 \mathrm{~m} / \mathrm{s}\) and (ii) \(v_{0}\) is \(9.5 \mathrm{~m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\max },\) no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\max }\). The value \(v_{\max }\) has a simple physical interpretation. What is it? (d) If \(u_{0}\) is less than some value \(v_{\min }\), no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\min }\). The value \(v_{\min }\) also has a simple physical interpretation. What is it?

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of \(20 \mathrm{~m} / \mathrm{s}\) (45 mi/h) when it reaches the end of the 120 -m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of \(20 \mathrm{~m} / \mathrm{s}\). What distance does the traffic travel while the car is moving the length of the ramp?

A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is \(5.0 \mathrm{~m}\) above the surface and has a downward speed of \(0.8 \mathrm{~m} / \mathrm{s}\). With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is \(1.6 \mathrm{~m} / \mathrm{s}^{2}\).

A block moving on a horizontal surface is at \(x=0\) when earth's surface. At \(1.15 \mathrm{~s}\) after liftoff, the rocket clears the top of its launch platform, \(63 \mathrm{~m}\) above the \(\mathrm{ground.}\) After an additional \(4.75 \mathrm{~s},\) it is \(1.00 \mathrm{~km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 s part of its flight and (b) the first \(5.90 \mathrm{~s}\) of its flight.
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