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Starting from rest, a boulder rolls down a hill with constant acceleration and travels 2.00 \(\mathrm{m}\) during the first second. (a) How far does it travel during the second second? (b) How fast is it moving at the end of the first second? at the end of the second second?

Short Answer

Expert verified
Distance during second second: 6 m; velocity at end of first second: 4 m/s; velocity at end of second second: 8 m/s.

Step by step solution

01

Understand the Problem

A boulder rolls down a hill starting from rest with constant acceleration. We need to find how far it travels during the second second and its velocity at the end of the first and second seconds. Given that it travels 2.00 m during the first second.
02

Establish Known Variables

From the problem, we know that the initial velocity \( u = 0 \) m/s, the distance traveled in the first second \( d_1 = 2.00 \) m, and time \( t_1 = 1 \) s. We want to find the acceleration \( a \) first, then find the distance traveled during the second second \( d_2 \), and the velocities at different times.
03

Calculate Acceleration

Using the kinematic equation for distance traveled, \( d = ut + \frac{1}{2}a t^2 \), we plug in the values for the first second: \[ 2.00 = 0 \cdot 1 + \frac{1}{2}a(1)^2 \] This simplifies to \( 2 = \frac{1}{2}a \). Solving for \( a \) gives \( a = 4 \) m/s².
04

Calculate Distance During Second Second

To find the distance traveled during the second second, calculate the total distance traveled in 2 seconds using the same formula: \[ d_2 = 0 \cdot 2 + \frac{1}{2}(4)(2)^2 = 8 \text{ m} \]Thus, the distance during the second second \( d_{2nd} = d_2 - d_1 = 8 - 2 = 6 \) m.
05

Calculate Velocity at the End of Each Second

Use the formula for final velocity \( v = u + at \).- At the end of the first second: \[ v_1 = 0 + 4 \cdot 1 = 4 \text{ m/s} \]- At the end of the second second: \[ v_2 = 0 + 4 \cdot 2 = 8 \text{ m/s} \]

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Key Concepts

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

Constant Acceleration
In physics, constant acceleration means that an object's velocity changes by the same amount every second. This simplification makes our calculations much more straightforward. When a boulder rolls down a hill under constant acceleration, it experiences a steady increase in speed, which provides a reliable foundation for solving kinematic problems.
For our exercise, the boulder's constant acceleration is 4 m/s² according to the calculation. This value, called 'magnitude of acceleration', tells us how much the boulder's speed increases each second. Since acceleration stays unchanged, we can use kinematic equations confidently to predict both how far the boulder travels and how fast it moves over time.
Think of constant acceleration as the boulder getting a consistent "push" that steadily speeds it up as it rolls downhill. It's essential for understanding motion and forms the basis for many introductory physics problems.
Displacement Calculation
Displacement in physics refers to how far an object moves in a specific direction. To calculate displacement when dealing with constant acceleration, we use the kinematic equation:
  • \( d = ut + \frac{1}{2}at^2 \)
where:
  • \( d \) is the displacement
  • \( u \) is the initial velocity
  • \( a \) is the acceleration
  • \( t \) is the time elapsing
It's important to apply this equation correctly. Start from rest (where initial velocity \( u = 0 \)) simplifies calculations. For example, to find the boulder's displacement during the first and second seconds:
  • First-second displacement is 2 m, calculated as shown before.
  • Total displacement for two seconds is determined to be 8 m.
  • Distance traveled during the second second is found by subtracting first-second displacement from the total, yielding 6 m.
Remember, displacement isn't just about total distance; it also includes direction.
Velocity Calculation
Calculating velocity is a fundamental aspect of analyzing motion under constant acceleration. Velocity measures how quickly an object's position changes. For constant acceleration, it changes uniformly, and we can use the formula:
  • \( v = u + at \)
where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is the acceleration
  • \( t \) is the time that has passed
This equation accounts for the constant "push" the boulder receives as it speeds up. Let's see its application throughout our example:
  • At the end of the first second, the boulder's velocity is 4 m/s.
  • At the end of the second second, it reaches a velocity of 8 m/s.
By understanding this equation, you can easily predict how fast an object moves at any point in time as long as you know the constants involved. Look at this change as a continuous increase in how far the boulder travels each second.

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Most popular questions from this chapter

According to recent typical test data, a Ford Focus travels 0.250 mi in 19.9 s starting from rest. The same car, when braking from 60.0 mph on dry pavement, stops in 146 ft. Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up. (a) Find this car's acceleration while braking and while speeding up. (b) If its acceleration is constant while speeding up, how fast (in mph) will the car be traveling after 0.250 mi of acceleration? (c) How long does it take the car to stop while braking from 60.0 \(\mathrm{mph} ?\)

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The "reaction time" of the average automobile driver is about 0.7 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 \(\mathrm{ft} / \mathrm{s}^{2}\) , compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 \(\mathrm{mi} / \mathrm{h}(\mathrm{in}\) a school zone) and (b) from an initial velocity of 55.0 \(\mathrm{mi} / \mathrm{h}\)

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