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Chapter 2: Problem 28

A car is approaching a hill at \(30.0 \mathrm{m} / \mathrm{s}\) when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of \(-2.00 \mathrm{m} / \mathrm{s}^{2}\) while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking \(x=0\) at the bottom of the hill, where \(v_{i}=30.0 \mathrm{m} / \mathrm{s} .\) (b) Determine the maximum distance the car rolls up the hill.

Short Answer

Expert verified
The position function is given by \(x(t) = 30t - t^{2}\) and the velocity function is \(v(t) = 30 - 2t\). The maximum distance the car moves up the hill is \(225 \mathrm{m}\).

Step by step solution

01

Express position and velocity functions

The generalized functions of motion are given as follows: \(x = x_i + v_i t + \frac{1}{2} a t^2\) for position and \(v = v_i + a t\) for velocity. Here, \(x_i\) is the initial position, \(v_i\) is the initial velocity, \(a\) is acceleration and \(t\) is time. For this problem, \(x_i = 0\), \(v_i = 30.0 \mathrm{m/s}\), and \(a= -2.00 \mathrm{m/s^{2}}\). Substituting these values, the functions become: \(x(t) = 30 t - t^2\) and \(v(t) = 30 - 2t\).
02

Determine when the car stops

The car stops moving up the hill when its velocity is zero. Thus, set the velocity function to zero and solve for time: \(0 = 30 - 2t\). Solving this gives \(t = 15 \mathrm{s}\).
03

Determine the maximum distance

Substitute \(t = 15 \mathrm{s}\) into the position function to get the maximum distance the car moves up the hill: \(x(15) = 30(15) - (15)^2 = 225 \mathrm{m}\).

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Key Concepts

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

Constant Acceleration
When a car's engine fails while going up a hill, its motion is influenced by constant acceleration. In this example, the car experiences a constant acceleration of \(-2.00 \text{ m/s}^2\). This means every second, the car鈥檚 velocity decreases by 2 m/s. Constant acceleration allows us to use specific equations, as the rate of change is predictable over time. This steady decrease makes calculations simpler and more straightforward.

Consider some key aspects of constant acceleration:
  • The magnitude indicates how quickly velocity changes. Positive values mean speeding up, while negative values (as in this scenario) show slowing down.
  • It's constant, meaning it doesn鈥檛 vary with time or position during the motion up the hill.
Understanding constant acceleration provides us with the foundation to analyze and solve motion problems effectively.
Motion Equations
Motion equations are tools that help us predict how an object moves over time. Two essential equations are used for position and velocity. These are rooted in physics and derived from the basic principles of motion.

For the position (\[x(t) = x_i + v_i t + \frac{1}{2} a t^2\]) and velocity functions (\[v(t) = v_i + a t\]), the interpretations are crucial:
  • \(x_i = 0\): This is the starting point at the hill鈥檚 base.
  • \(v_i = 30 \text{ m/s}\): The initial speed as the car begins its ascent.
  • \(a = -2.00 \text{ m/s}^2\): Constant acceleration slowing the car.
When these values are plugged into the equations, we get:
  • Position: \(x(t) = 30t - t^2\)
  • Velocity: \(v(t) = 30 - 2t\)
With these equations, you can find how far the car travels and its speed at any moment during its uphill journey.
Maximum Distance
Finding the maximum distance involves understanding when the car up the hill comes to a complete stop.

This occurs when its velocity reaches zero. By using the velocity equation \(v(t) = 30 - 2t\) and setting it to zero (\[30 - 2t = 0\]), we solve for time \(t = 15 \text{ seconds}\).

Next, substitute \(t = 15\) into the position equation \(x(t) = 30t - t^2\) to compute the maximum distance:
  • \(x(15) = 30 \times 15 - (15)^2 = 225 \text{ meters}\)
The car rolls up 225 meters before stopping. Calculating maximum distance is essential in understanding the limits of motion under constant acceleration.

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Most popular questions from this chapter

It is possible to shoot an arrow at a speed as high as \(100 \mathrm{m} / \mathrm{s} .\) (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?

Astronauts on a distant planet toss a rock into the air. With the aid of a camera that takes pictures at a steady rate, they record the height of the rock as a function of time as given in Table \(\mathrm{P} 2.74\). (a) Find the average velocity of the rock in the time interval between each measurement and the next. (b) Using these average velocities to approximate instantaneous velocities at the midpoints of the time intervals, make a graph of velocity as a function of time. Does the rock move with constant acceleration? If so, plot a straight line of best fit on the graph and calculate its slope to find the acceleration.$$\begin{array}{llll} \hline \text { Time (s) } & \text { Height }(\mathbf{m}) & \text { Time }(\mathbf{s}) & \text { Height }(\mathbf{m}) \\\\\hline \text { Times } & & \\\\\hline 0.00 & 5.00 & 2.75 & 7.62 \\\0.25 & 5.75 & 3.00 & 7.25 \\ 0.50 & 6.40 & 3.25 & 6.77 \\\0.75 & 6.94 & 3.50 & 6.20 \\\1.00 & 7.38 & 3.75 & 5.52 \\\1.25 & 7.72 & 4.00 & 4.73 \\\1.50 & 7.96 & 4.25 & 3.85 \\\1.75 & 8.10 & 4.50 & 2.86 \\\2.00 & 8.13 & 4.75 & 1.77 \\\2.25 & 8.07 & 5.00 & 0.58 \\\2.50 & 7.90 & & \\\\\hline\end{array}$$.

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by \(a=-3.00 v^{2}\) for \(v>0 .\) If the marble enters this fluid with a speed of \(1.50 \mathrm{m} / \mathrm{s}\), how long will it take before the marble's speed is reduced to half of its initial value?

The position of a particle moving along the \(x\) axis varies in time according to the expression \(x=3 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. Evaluate its position (a) at \(t=3.00 \mathrm{s}\) and (b) at \(3.00 \mathrm{s}+\Delta t\). (c) Evaluate the limit of \(\Delta x / \Delta t\) as \(\Delta t\) approaches zero, to find the velocity at \(t=3.00 \mathrm{s}\).

An electron in a cathode ray tube (CRT) accelerates from \(2.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) to \(6.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) over \(1.50 \mathrm{cm} .\) (a) How long does the electron take to travel this \(1.50 \mathrm{cm} ?\).(b) What is its acceleration?
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