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A ball starts from rest and rolls down a hill with uniform acceleration, traveling 200 m during the second 5.0 s of its motion. How far did it roll during the first 5.0 s of motion?

Short Answer

Expert verified
The ball rolled 66.67 meters during the first 5 seconds.

Step by step solution

01

Understanding the Problem

A ball rolls down a hill with uniform acceleration. It covers 200 meters during the second 5 seconds of its motion. We need to find out how far it rolls during the first 5 seconds of its motion.
02

Identify Known Variables

The known quantities are the distance traveled during the second 5 seconds, which is 200 m, and the time period of interest being the first 5 seconds.
03

Apply the Uniform Acceleration Formula

For an object starting from rest (initial velocity, \( u = 0 \)), the distance \( s \) covered in time \( t \) under uniform acceleration \( a \) is given by the equation: \( s = ut + \frac{1}{2}at^2 \). Simplifying since \( u = 0 \), we have \( s = \frac{1}{2}at^2 \).
04

Find Distance Covered in First 10 Seconds

The object moves a distance \( s_1 \) in 5 seconds and \( s_2 \) between 5 and 10 seconds. The difference \( s_2 - s_1 = 200 \) m. The total distance after 10 seconds is \( \frac{1}{2}a(10)^2 \). Therefore: \( s_1 = \frac{1}{2}a(5)^2 \) and \( 200 = \frac{1}{2}a(10)^2 - \frac{1}{2}a(5)^2 \).
05

Solve for Acceleration

Solve the equation: \( 200 = \frac{1}{2}a(10^2 - 5^2) \). Simplifying gives \( 200 = \frac{1}{2}a(100 - 25) \) or \( 200 = \frac{1}{2}a(75) \). Thus, \( a = \frac{400}{75} \).
06

Calculate Distance in First 5 Seconds

Now using \( s_1 = \frac{1}{2}a(5)^2 \) and substituting the value of \( a \), we find \( s_1 = \frac{1}{2} \times \frac{400}{75} \times 25 \). This simplifies to \( s_1 = 66.67 \) m.

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Key Concepts

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

Uniform Acceleration
Uniform acceleration occurs when an object speeds up or slows down at a constant rate over time. This is an important concept in kinematics because it allows us to predict the future motion of objects when their acceleration doesn't change.
In a problem involving uniform acceleration like the exercise provided, we often use the formula:
  • For an object starting from rest: \[ s = \frac{1}{2} a t^2 \] where:
    • \( s \) is the distance traveled,
    • \( a \) is the uniform acceleration,
    • \( t \) is the time elapsed.
Understanding and applying this formula is crucial for solving problems where the speed of the object is either increasing or decreasing at a constant rate. The key takeaway is to determine the acceleration before calculating how far an object travels in a given time.
Distance Traveled
In the context of physics and kinematics, the distance traveled by an object is the length of the path it takes during its motion. For objects moving with uniform acceleration, determining the distance involves knowing both the acceleration rate and the time period of interest.
In our given exercise, we are asked to find out the distance a ball rolls during the first 5 seconds under uniform acceleration. Compatibility with the formula for uniform acceleration means that once we compute the acceleration (using the data for the second 5-second interval), we can plug it back into our formula to find the distance for any other time frame.
Applying the formula \[ s_1 = \frac{1}{2} a t^2 \] for the first 5 seconds gives us the specific distance the ball traveled as it started from rest. Understanding how to isolate and plug-in known values helps solve these types of physics questions effectively.
Physics Problem Solving
Physics problem solving involves systematic steps, starting with understanding the problem and identifying known variables.
In our exercise, solving involved:
  • Identifying variables like distance traveled during a defined time interval and knowing the object's start condition (rest in this case).
  • Applying relevant formulas: Here, the formula for uniformly accelerated motion for an object starting from rest was applied.
  • Algebraic manipulation: Equating known distances helped isolate unknown variables like acceleration.
  • Iterative calculations: Using solved acceleration to then determine traveled distance in another interval.
The clarity in each step ensures correct setup of equations and helps avoid errors based on faulty initial assumptions or calculations.
This structured approach is crucial for success in both educational exercises and real-world physics applications.
Motion Dynamics
Motion dynamics delve into how an object's movement changes under the influence of forces. In kinematics, specifically when dealing with uniform acceleration, motion is typically analyzed in a linear path where forces like gravity are constant, such as our example with the rolling ball on a hill.

The constant influence of gravity accelerates the ball uniformly, and inherently affects how the calculations for distance and acceleration need to be conducted. Dynamics help explain the cause and effect relationship between applied forces and motion, emphasizing the need for understanding fundamentals like Newton's laws in more complex scenarios.
In simpler uniform motion scenarios, these dynamics help us derive formulas and principles like those used in our given problem set, making it possible to predict future movements of objects accurately.

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Most popular questions from this chapter

A jet fighter pilot wishes to accelerate from rest at a constant acceleration of 5\(g\) to reach Mach 3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.0 s. Use 331 m/s for the speed of sound. (a) Will the period of acceleration last long enough to cause him to black out? (b) What is the greatest speed he can reach with an acceleration of 5\(g\) before he blacks out?

Cars \(A\) and \(B\) travel in a straight line. The distance of \(A\) from the starting point is given as a function of time by \(x_A(t) = \alpha{t} + \beta{t}^2\), with \(\alpha =\) 2.60 m/s and \(\beta =\) 1.20 m/s\(^2\). The distance of \(B\) from the starting point is \(x_B(t) = \gamma{t}^2 - \delta{t}^3\), with \(\gamma =\) 2.80 m/s\(^2\) and \(\delta =\) 0.20 m/s\(^3\). (a) Which car is ahead just after the two cars leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 m/s relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?

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 s later. Ignore air resistance. (a) If the height of the building is 20.0 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 m/s and (ii) \(v_0\) is 9.5 m/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 \(v_0\) 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?

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

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