/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A car sits on an entrance ramp t... [FREE SOLUTION] | 91Ó°ÊÓ

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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 m/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 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Short Answer

Expert verified
(a) \( a = \frac{5}{3} \, \text{m/s}^2 \), (b) \( t = 12 \, \text{s} \), (c) \( d = 240 \, \text{m} \).

Step by step solution

01

Identify Known and Unknowns for Part (a)

The car starts from rest, which means its initial velocity \( u = 0 \). The final velocity \( v = 20 \, \text{m/s} \), and the distance \( s = 120 \, \text{m} \). We need to find the acceleration \( a \).
02

Use Equation of Motion for Acceleration

Using the equation \( v^2 = u^2 + 2as \), substitute the known values: \( (20)^2 = (0)^2 + 2a(120) \). Solving for \( a \), we get \( a = \frac{400}{240} = \frac{5}{3} \, \text{m/s}^2 \).
03

Identify Known and Unknowns for Part (b)

Now, we need to find the time \( t \) it takes to travel the ramp. We know \( u = 0 \), \( v = 20 \, \text{m/s} \), and \( a = \frac{5}{3} \, \text{m/s}^2 \).
04

Use Equation of Motion for Time

Using the equation \( v = u + at \), substitute the known values: \( 20 = 0 + \frac{5}{3}t \). Solving for \( t \), we find \( t = 12 \, \text{s} \).
05

Identify Known and Unknowns for Part (c)

The traffic on the freeway is moving at a constant speed of \( 20 \, \text{m/s} \). We need to determine the distance traveled by the traffic in 12 seconds.
06

Calculate Distance Traveled by Traffic

Since distance \( d = vt \) where \( v = 20 \, \text{m/s} \) and \( t = 12 \, \text{s} \), the distance \( d = 20 \times 12 = 240 \, \text{m} \).

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Key Concepts

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

constant acceleration
Constant acceleration means that the rate at which the speed or velocity of an object changes remains steady over time. In simpler terms, it means the object gains the same amount of speed every second. Imagine you are pressing down steadily on the pedal of a car. If you're accelerating at a constant rate, your speed increases smoothly, without any sudden spikes or drops.
Constant acceleration is an essential concept to understand because it simplifies the calculations of motion. When acceleration is constant:
  • The behaviors of velocity and displacement become predictable.
  • Motions can be described using basic equations.
Understanding constant acceleration helps in solving problems like the one in our exercise, where the car accelerates briskly up to freeway speed.
equations of motion
Equations of motion are vital tools in kinematics, the study of movement. These equations describe the mathematical relationship between displacement, velocity, acceleration, and time. For objects moving with constant acceleration, we have three primary equations:
  • The first is: \( v = u + at \)
    It connects current velocity \( v \), initial velocity \( u \), acceleration \( a \), and time \( t \).
  • The second is: \( s = ut + \frac{1}{2}at^2 \)
    This relates displacement \( s \) with initial velocity, time, and acceleration.
  • The third is: \( v^2 = u^2 + 2as \)
    It shows the connection between velocities, acceleration, and displacement without involving time.
These equations allow you to solve for unknown motion values based on what is known. In our exercise, they help calculate the car’s acceleration and the time it takes to travel the ramp.
velocity and time
Velocity tells us how fast something is moving and in what direction. It's a vector quantity, meaning it includes both speed and direction. In the concept of motion under constant acceleration, velocity changes steadily.
When solving problems in kinematics, it's crucial to know either the initial or final velocity, along with the time, to find other unknowns. For example, using the equation \( v = u + at \), if the car starts from rest (\( u = 0 \)) and accelerates to \( 20 \, \text{m/s} \) over some time, we can determine both the acceleration and the time elapsed. Time is a fundamental factor:
  • It's essential for calculating how long an object takes to get from one point to another.
  • It also helps predict future motion states.
In our problem, understanding the timing helped determine how long the ramp journey took and calculated distances.
distance traveled
Distance traveled is a measure of the total length of the path taken by an object in motion. In straight-line motion with constant acceleration, calculating distance can be straightforward using motion equations.
The distance covered can be calculated using the expression:
\[ s = ut + \frac{1}{2}at^2 \]
Here, \( u \) is the initial velocity, \( a \) is acceleration, and \( t \) is the time duration of travel. In scenarios where speed is constant, the equation simplifies to:
\[ d = vt \]
This is shown in our problem, where the steady-speed traffic on the freeway covers double the length of the ramp as the accelerating car, simply using its constant speed over time.
  • Understanding how to compute distance provides insight into the motion and confirms other calculations.
  • It's a vital concept when determining changes in an object's position over time.
By mastering the distance traveled, you gain a deeper grasp of how motion translates into real-world paths.

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Most popular questions from this chapter

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 m/s\(^2\) upward. At 25.0 s after launch, the second stage fires for 10.0 s, which boosts the rocket's velocity to 132.5 m/s upward at 35.0 s after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stage-two rocket be moving just as it reaches the launch pad?

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (\(\textbf{Fig. E2.44}\)). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_y-t\), \(v_y-t\), and \(y-t\) graphs for the motion.

A bird is flying due east. Its distance from a tall building is given by \(x(t) =\) 28.0 m \(+\) (12.4 m/s)\(t\) - (0.0450 m/s\(^3)t^3\). What is the instantaneous velocity of the bird when \(t =\) 8.00 s?

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

A turtle crawls along a straight line, which we will call the \(x\)-axis with the positive direction to the right. The equation for the turtle's position as a function of time is \(x(t) =\) 50.0 cm + (2.00 cm/s)\(t -\) (0.0625 cm/s\(^2)t^2\). (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 cm from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of \(x\) versus \(t, v_x\) versus \(t\), and \(a_x\) versus \(t\), for the time interval \(t =\) 0 to \(t =\) 40 s.

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