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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 it 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 200 \(\mathrm{s}\) 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 \(2.00-\mathrm{s}\) 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 m point? (e) How far did the sled go during the first second after passing the 14.4 -m point?

Step by step solution

01

Calculate average velocity for each interval

The average velocity is calculated using the formula: \[ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} \]For the interval from 14.4 m to 25.6 m:\[ \text{Average Velocity} = \frac{25.6 \, \text{m} - 14.4 \, \text{m}}{2.00 \, \text{s}} = \frac{11.2 \, \text{m}}{2.00 \, \text{s}} = 5.6 \, \text{m/s} \]For the interval from 25.6 m to 40.0 m:\[ \text{Average Velocity} = \frac{40.0 \, \text{m} - 25.6 \, \text{m}}{2.00 \, \text{s}} = \frac{14.4 \, \text{m}}{2.00 \, \text{s}} = 7.2 \, \text{m/s} \]For the interval from 40.0 m to 57.6 m:\[ \text{Average Velocity} = \frac{57.6 \, \text{m} - 40.0 \, \text{m}}{2.00 \, \text{s}} = \frac{17.6 \, \text{m}}{2.00 \, \text{s}} = 8.8 \, \text{m/s} \]

02

Use acceleration to find increment per interval

Since the sled undergoes constant acceleration, the increase in velocity can be determined by examining the differences in average velocity: - Between the first and second intervals: \[ 7.2 \, \text{m/s} - 5.6 \, \text{m/s} = 1.6 \, \text{m/s}\]- Between the second and third intervals: \[ 8.8 \, \text{m/s} - 7.2 \, \text{m/s} = 1.6 \, \text{m/s}\]This constant increase in velocity per interval (i.e., every 2 seconds) indicates an acceleration of:\[ a = \frac{\Delta v}{\Delta t} = \frac{1.6 \, \text{m/s}}{2 \, \text{s}} = 0.8 \, \text{m/s}^2 \]

03

Calculate initial speed at 14.4 m point

We can deduce the initial velocity when the sled is at the 14.4 m point by considering the average velocity over the first interval. Since the average velocity (5.6 m/s) is the midpoint between the initial velocity \(v_0\) and the final velocity \(v_f\):\[ \text{Average Velocity} = \frac{v_0 + v_f}{2} = 5.6 \, \text{m/s} \]Setting: \[ v_f = v_0 + a \times 2 \] \[ 5.6 = \frac{v_0 + (v_0 + 1.6)}{2} \Rightarrow v_0 = 4.8 \, \text{m/s} \]

04

Calculate time to reach 14.4 m

Using the kinematic equation:\[ s = v_0 t + \frac{1}{2} a t^2 \]where \(s = 14.4\) m, \(v_0 = 0\) m/s (since it starts from rest), and \(a = 0.8\, \text{m/s}^2\):\[ 14.4 = 0 + \frac{1}{2} \cdot 0.8 \cdot t^2 \]\[ 14.4 = 0.4 t^2 \]\[ t^2 = \frac{14.4}{0.4} = 36 \]\[ t = \sqrt{36} = 6 \, \text{s} \]

05

Calculate distance traveled in first second after 14.4 m

We can calculate the distance covered in the first second after passing 14.4 m using the formula:\[ s = v_0 t + \frac{1}{2} a t^2 \]where \(v_0 = 4.8\, \text{m/s}\), \(t = 1\, \text{s}\), and \(a = 0.8\, \text{m/s}^2\):\[ s = 4.8 \times 1 + \frac{1}{2} \times 0.8 \times 1^2 \]\[ s = 4.8 + 0.4 = 5.2 \, \text{m} \]

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Key Concepts

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

Constant Acceleration

In the world of kinematics, constant acceleration describes a situation where an object's speed increases or decreases at a uniform rate over time. This means the object is steadily gaining or losing velocity in a controlled fashion. When the sled slides down the hill, it experiences constant acceleration due to gravitational pull, assuming no other forces (like friction) are acting significantly. This simplifies calculations, as we use consistent values for acceleration, like the one solved in the exercise, at 0.8 m/s². Constant acceleration allows us to predict future positions and velocities using straightforward equations.

Average Velocity

Average velocity gives us a snapshot of how fast an object is moving over a specified interval. It is calculated by dividing the total displacement by the time taken for that displacement. Consider the sled moving from 14.4 m to 25.6 m; its average velocity in this interval is 5.6 m/s as calculated by

• Displacement: 11.2 m (25.6 m - 14.4 m)
• Time: 2.00 s

Average velocity helps in understanding movement over intervals and isn't affected by changes within that interval. Instead, it tells us the constant speed needed to cover the same distance in the same time.

Kinematic Equations

Kinematic equations are integral tools in physics that describe motion with constant acceleration. They allow us to find unknown properties of motion like distance, time, acceleration, and velocity. The equations are especially useful when looking at problems like a sled sliding down a hill. These include patterns of motion which can be predicted using equations such as \[ s = v_0 t + \frac{1}{2} a t^2 \] where:

• \( s \) is the position
• \( v_0 \) is the initial velocity
• \( a \) is the acceleration
• \( t \) is the time

This formula, for instance, determines how far an object travels over a certain time, given its start speed and acceleration.

Distance-Time Calculations

Distance-time calculations are central to understanding motion. By relating velocity, time, and displacement, we can uncover how far and how fast an object has traveled over specific periods. Considering the sled's journey down the hill, we calculate its average velocity in stages. Given the calculated acceleration at 0.8 m/s², we utilize kinematic equations to find the time taken to reach points or the distance covered in new intervals—a vital aspect of predicting motion. These calculations provide insights into an object's trajectory, helping us understand not just movement, but the underlying forces and timing.