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A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time sled is 14.4 \(\mathrm{m}\) from the top, 2.00 s after that it is 25.6 \(\mathrm{m}\) from the top, 2.00 s later 40.0 \(\mathrm{m}\) from the top, and 2.00 s later it is 57.6 \(\mathrm{m}\) from the top. (a) What is the magnitude of the average velocity of the sled during each of the 2.00 -s intervals after passing the \(14.4-\mathrm{m}\) point? (b) What is the acceleration of the sled? (c) What is the speed of the sled when it passes the \(14.4-\mathrm{m}\) point? (d) How much time did it take to go from the top to the \(14.4-\mathrm{m}\) point? (e) How far did the sled go during the first second after passing the \(14.4-\mathrm{m}\) point?

Step by step solution

01

Calculate Average Velocity for Each Interval

For each 2-second interval, use the formula for average velocity, \( v_{avg} = \frac{\Delta s}{\Delta t} \), where \( \Delta s \) is the change in position, and \( \Delta t = 2 \ seconds \).1. First interval (from 14.4 m to 25.6 m): \[ v_{avg} = \frac{25.6 \ m - 14.4 \ m}{2 \ s} = \frac{11.2 \ m}{2 \ s} = 5.6 \ m/s \]2. Second interval (from 25.6 m to 40.0 m): \[ v_{avg} = \frac{40.0 \ m - 25.6 \ m}{2 \ s} = \frac{14.4 \ m}{2 \ s} = 7.2 \ m/s \]3. Third interval (from 40.0 m to 57.6 m): \[ v_{avg} = \frac{57.6 \ m - 40.0 \ m}{2 \ s} = \frac{17.6 \ m}{2 \ s} = 8.8 \ m/s \]

02

Calculate the Acceleration

To find the constant acceleration, we use the kinematic equations. Use the change in average velocity over time. \[ a = \frac{v_{f} - v_{i}}{t} \]Take any two consecutive intervals to find acceleration:\[ a = \frac{7.2 \ m/s - 5.6 \ m/s}{2 \ s} = \frac{1.6 \ m/s}{2 \ s} = 0.8 \ m/s^2 \]Check using the next pair to confirm consistency:\[ a = \frac{8.8 \ m/s - 7.2 \ m/s}{2 \ s} = \frac{1.6 \ m/s}{2 \ s} = 0.8 \ m/s^2 \]

03

Determine Speed at 14.4 m Point

Use the kinematic equation: \( v_{f}^2 = v_{i}^2 + 2a\Delta s \). Since the sled starts from rest, \( v_{i} = 0 \).\[ v_{f}^2 = 0 + 2(0.8 \ m/s^2)(14.4 \ m) \]\[ v_{f} = \sqrt{2(0.8)(14.4)} = \sqrt{23.04} = 4.8 \ m/s \]

04

Find Time to Reach 14.4 m Point

Use the kinematic formula for distance: \( s = v_{i}t + \frac{1}{2}at^2 \).Since the sled starts from rest, \( v_{i} = 0 \), thus:\[ 14.4 = \frac{1}{2}(0.8)t^2 \]\[ t^2 = \frac{14.4}{0.4} = 36 \]\[ t = \sqrt{36} = 6 \ seconds \]

05

Calculate Distance Traveled in First Second After 14.4 m

Use the formula for the distance traveled in the first second:\[ s = v_{i}t + \frac{1}{2}at^2 \]Here, \( v_{i} = 4.8 \ m/s \) (speed at 14.4 m), \( t = 1 \ s \), and \( a = 0.8 \ m/s^2 \):\[ s = (4.8)(1) + \frac{1}{2}(0.8)(1)^2 \]\[ s = 4.8 + 0.4 = 5.2 \ m \]

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Key Concepts

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

Average Velocity

Average velocity is a key concept in kinematics, which helps us understand the motion of objects. It is calculated as the change in position (or displacement) divided by the change in time.

• The formula for average velocity is given by: \( v_{\text{avg}} = \frac{\Delta s}{\Delta t} \), where \( \Delta s \) is the change in position, and \( \Delta t \) is the change in time.
• In simple terms, average velocity tells us how fast something is moving on average over a specific time period.
• For example, in the exercise provided, during the first interval after the sled passes the 14.4-meter point, it had an average velocity of 5.6 m/s over a 2-second span.

Understanding average velocity helps in breaking down motion into manageable pieces, making it easier to analyze each segment of an object's journey.

Constant Acceleration

Constant acceleration occurs when an object's velocity changes at a uniform rate over time.

• This means that every second, the object's velocity increases or decreases by the same amount.
• In the given problem, the sled experienced constant acceleration as it moved down the hill.
• We can calculate constant acceleration using the equation: \[ a = \frac{v_{f} - v_{i}}{t} \]

Here, \( v_{f} \) and \( v_{i} \) represent the final and initial velocities respectively, while \( t \) is the time over which this change happens. Knowing constant acceleration allows us to predict how an object's velocity will evolve over time.

Kinematic Equations

Kinematic equations are essential tools in physics that describe motion in a straight line under constant acceleration. They tie together time, velocity, acceleration, and displacement.

• These equations are powerful because they let us calculate vast amounts of information once we know a few key variables.
  • An example from the problem is the equation: \[ v_{f}^2 = v_{i}^2 + 2a\Delta s \] which helps us determine the final speed based on initial speed, acceleration, and displacement.
• Another useful kinematic formula is: \[ s = v_{i}t + \frac{1}{2}a t^2 \] This helps calculate the position of an object at any given time "t".

Mastering these equations gives us powerful insights into understanding how objects move.

Distance Traveled

Distance traveled, in kinematics, refers to the total length of the path taken by an object. It's crucial to differentiate between distance and displacement.

• Displacement is the direct "straight line" distance between starting and ending points, while distance traveled is the sum of all distances covered.
• In the example exercise, determining the distance traveled involved calculating how far the sled moved between specified points, such as from 14.4 meters to 25.6 meters.
• To find the distance for time intervals where acceleration and initial velocity are known, the formula \[ s = v_{i}t + \frac{1}{2}a t^2 \] is very helpful.

Understanding the distance traveled not only gives us clarity on the path taken but also assists in understanding the complete picture of an object's motion.