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Chapter 2: Problem 24

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?

Short Answer

Expert verified
The acceleration of the car is \(1.67 \mathrm{~m/s}^2\), the time it takes for the car to travel the length of the ramp is \(12.0 \mathrm{~s}\), and the distance the traffic travels while the car moves the ramp length is \(240 \mathrm{~m}\).

Step by step solution

01

Find the acceleration of the car

We know the car starts from rest, moves in a straight line, and attains a speed of \(20 \mathrm{~m} / \mathrm{s}\) after travelling on the 120-m-long ramp. Given these conditions, one appropriate kinematic equation is \(v^2 = u^2 + 2a(x-x_0)\), where \(v\) is the final velocity, \(u\) is initial velocity, \(a\) is acceleration and \((x-x_0)\) is the displacement. We know that \(u = 0\), \(v = 20 \mathrm{~m} / \mathrm{s}\) and \((x-x_0) = 120 \mathrm{~m}\). Substituting these values into the equation and solving for \(a\), we get \(a = \frac{v^2 - u^2}{2(x - x_0)} = \frac{20^2 - 0^2}{2 (120)} = 1.67 \mathrm{~m/s}^2\).
02

Calculate how much time it takes for the car to travel the length of the ramp

Now we know the acceleration and we can use this value to find the time. We use the kinematic equation \(v = u + at\), where \(v = 20 \mathrm{~m} / \mathrm{s}\), \(u = 0 \mathrm{~m} / \mathrm{s}\), \(a = 1.67 \mathrm{~m/s}^2\) and \(t\) is the time which we want to find. Solving for \(t\), we get \(t = \frac{v - u}{a} = \frac{20}{1.67} = 12.0 \mathrm{~s}\).
03

Determine the distance the traffic travels while the car is moving the length of the ramp

We can find the distance the traffic travels during this time duration by using the formula \(d = vt\), where \(d\) is distance, \(v\) is velocity (which for the traffic is \(20 \mathrm{~m} / \mathrm{s}\)) and \(t = 12.0 \mathrm{~s}\). So, substituting into the formula, we have \(d = 20 \times 12.0 = 240 \mathrm{~m}\).

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Key Concepts

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

Constant Acceleration
Constant acceleration is a crucial concept in kinematics. It refers to the scenario where the rate of change of velocity of an object remains consistent over time. When a car on a ramp accelerates at a constant rate from rest to a certain speed, its increase in velocity is uniform throughout.
To solve problems involving constant acceleration, you often rely on kinematic equations, which help connect various motion parameters such as displacement, velocity, and time.
  • In our exercise, the car's acceleration is constant, meaning it does not change as the car travels along the ramp. This allows us to use straightforward formulas without delving into variable accelerations.
  • Understanding constant acceleration simplifies calculations and predictions about the object's motion, as future positions and speeds are easily derived from the consistent acceleration pattern.
Kinematic Equations
Kinematic equations are mathematical expressions that describe the motion of objects without considering the forces that cause the motion. They are particularly useful when dealing with constant acceleration.
In our exercise, we use two key kinematic equations:
  • The first equation: \[v^2 = u^2 + 2a(x-x_0)\]helps us determine the acceleration of the car. By knowing the final velocity \(v\), initial velocity \(u\) (which is zero since it starts from rest), and the displacement \((x-x_0)\), we can solve for acceleration \(a\).
  • The second equation: \[v = u + at\]allows us to calculate the time \(t\) taken for the car to reach the end of the ramp. Once the acceleration is known, finding the time is straightforward.
Using these equations requires knowing and substituting the right values for each variable. The simplicity and versatility of kinematic equations make them indispensable in solving physics problems involving motion.
Velocity-Time Relationship
The velocity-time relationship is a fundamental aspect of understanding an object's motion, especially under constant acceleration. This relationship describes how velocity changes with time.
For the car in our exercise, which accelerates from rest to a specific speed, the velocity-time graph would show a straight line, indicating constant acceleration.
  • The initial point on the graph, where velocity is zero, represents the start of motion from rest.
  • As time progresses, the velocity increases linearly, showcasing that the acceleration is steady and unchanging.
  • The slope of this line represents the acceleration rate. Hence, a steeper slope indicates a higher acceleration.
  • Finally, by calculating the area under the velocity-time graph, we can determine the distance traveled, reinforcing the connection between velocity, time, and displacement.
Understanding the velocity-time relationship helps predict how an object's speed will change, providing insights into both its past and future motion under constant acceleration.

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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?

If the aorta (diameter \(d_{\mathrm{a}}\) ) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? (a) \(\sqrt{d_{a}} ;\) (b) \(d_{\mathrm{a}} / \sqrt{2} ;\) (c) \(2 d_{\mathrm{a}} ;\) (d) \(d_{\mathrm{a}} / 2\)

A 15 kg rock is dropped from rest on the earth and reaches the ground in \(1.75 \mathrm{~s}\). When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in \(18.6 \mathrm{~s}\). What is the acceleration due to gravity on Enceladus?

An antelope moving with constant acceleration covers the distance between two points \(70.0 \mathrm{~m}\) apart in \(6.00 \mathrm{~s}\). Its speed as it passes the second point is \(15.0 \mathrm{~m} / \mathrm{s}\). What are (a) its speed at the first point and (b) its acceleration?

A sprinter runs a \(100 \mathrm{~m}\) dash in \(12.0 \mathrm{~s}\). She starts from rest with a constant acceleration \(a_{x}\) for \(3.0 \mathrm{~s}\) and then runs with constant speed for the remainder of the race. What is the value of \(a_{x} ?\)
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