/*! 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 69 A ball starts from rest and roll... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 \(\mathrm{m}\) during the second 5.0 \(\mathrm{s}\) of its motion. How far did it roll during the first 5.0 s of motion?

Short Answer

Expert verified
The ball rolled 50 meters during the first 5 seconds of motion.

Step by step solution

01

Understand the Problem

We are given a ball rolling down a hill with uniform acceleration. It travels 150 m during the second 5.0 seconds. We need to find the distance it rolled during the first 5.0 seconds.
02

Use the Equations of Motion

Let's use the equations of motion, specifically: \[ s = ut + \frac{1}{2}at^2 \]where:\( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
03

Set Initial Conditions for First Part

For the first 5.0 seconds, the ball starts from rest, so \( u = 0 \). Thus, the distance \( s_1 \) for the first 5.0 seconds is:\[ s_1 = \frac{1}{2}a(5)^2 = \frac{25}{2}a \]
04

Calculate Distance for the First 10 Seconds

Since we know the distance traveled during the second 5 seconds (from 5 s to 10 s) is 150 m, we calculate the total distance \( s_{10} \) traveled in the first 10 seconds:\[ s_{10} = \frac{1}{2}a(10)^2 = 50a \]
05

Calculate the Magnitude of the Acceleration

The distance traveled during the second 5 seconds \( (s_{10} - s_1) \) is 150 meters. Therefore:\[ s_{10} - s_1 = 150 \]Substituting previous expressions, we have:\[ 50a - \frac{25}{2}a = 150 \]
06

Solve for Acceleration

Solve the equation from Step 5:\[ 50a - 12.5a = 150 \]\[ 37.5a = 150 \]\[ a = 4 \text{ m/s}^2 \]
07

Calculate Distance for the First 5 Seconds

Use the value of acceleration to calculate \( s_1 \):\[ s_1 = \frac{25}{2} \times 4 = 50 \text{ m} \]

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Key Concepts

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

Uniform Acceleration
When we talk about uniform acceleration in physics, we're referring to a situation where an object speeds up or slows down at a constant rate. This means the acceleration does not change over time. In our exercise, the ball rolls down a hill under the influence of gravity, experiencing uniform acceleration throughout its motion. This makes things a bit simpler because the equations we use to describe the motion can take advantage of this constancy.
Uniform acceleration is a key concept when handling problems where the speed of an object increases or decreases consistently. It allows us to predict how far and fast an object will travel over a specific period without needing to know all the intermediate values of speed. Such situations are common in things like a car accelerating from a stoplight or, in this case, a ball rolling down a hill. The predictability here is useful for solving the exercise, where knowing the acceleration lets us find out how far the ball has rolled at various times.
Equations of Motion
Equations of motion are a set of formulas that describe the mathematical relationship between displacement, velocity, acceleration, and time. In kinematics, these equations are crucial in solving problems involving objects in motion. They allow us to find unknown values such as final velocity or distance traveled, given certain variables.
For our rolling ball, we use the equation \( s = ut + \frac{1}{2}at^2 \), which lets us find the distance covered when we know the initial velocity \( u \), the acceleration \( a \), and the time \( t \). Because the ball starts from rest, \( u = 0 \), simplifying our task. Thus, we only need to focus on the acceleration part of the equation. Using these equations, we effectively calculated the distance traveled by the ball during both the first and second intervals.
Initial Velocity
Initial velocity is a key element in physics problems, describing the speed and direction of an object when it first starts moving. It forms the basis of what happens in any motion thereafter. In our problem, since the ball is starting from rest, its initial velocity \( u \) is 0. This simplifies the calculations significantly because it removes the need to account for any initial speed in the motion equations.
By knowing the initial velocity, we can apply it to determine how acceleration affects the subsequent motion. This makes calculations more straightforward when using the equations of motion. When an object starts from rest, just like the ball rolling down the hill, focusing on how the acceleration contributes to the distance traveled becomes the primary task.

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Most popular questions from this chapter

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

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 \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2}\) (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 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}.\)

The position of the front bumper of a test car under microprocessor control is given by \(x(t)=2.17 \mathrm{m}+\) \(\left(4.80 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(0.100 \mathrm{m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x\) -t, \(v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{s}.\)

Catching the Bus. A student is running at her top speed of 5.0 \(\mathrm{m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still 40.0 \(\mathrm{m}\) from the bus, it starts to pull away, moving with a constant acceleration of 0.170 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) For how much time and what distance does the student have to run at 5.0 \(\mathrm{m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x\) -t graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 \(\mathrm{m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

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, it reaches the ground in 18.6 \(\mathrm{s}.\) What is the acceleration due to gravity on Enceladus?

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