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Bill steps off a 3.0 -m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of \(4.2 \mathrm{m} / \mathrm{s}\) from a \(1.0-\mathrm{m}\) -high diving board. Choosing the origin to be at the water's surface, and upward to be the positive \(x\) direction, write \(x\) -versus-t equations of motion for both Bill and Ted.

Short Answer

Expert verified
The x-t equations are: Bill: \(x_B(t) = 3.0 - 4.9t^2\); Ted: \(x_T(t) = 1.0 + 4.2t - 4.9t^2\).

Step by step solution

01

Understanding the scenario

We have two scenarios here: Bill stepping off a 3.0 m high diving board with an initial velocity of 0 m/s, and Ted jumping up from a 1.0 m high diving board with an initial velocity of 4.2 m/s. We will write equations of motion for each.
02

Set up the equation for Bill

For Bill, using the equation of motion: \[ x_B(t) = x_{B0} + v_{B0}t - \frac{1}{2}gt^2 \] where \(x_{B0} = 3.0 \) m (the height of the board), \(v_{B0} = 0 \) m/s (initial velocity), and \(g = 9.8 \) m/s² (acceleration due to gravity, downwards). The equation becomes: \[ x_B(t) = 3.0 - \frac{1}{2} \times 9.8 \times t^2 \] \[ x_B(t) = 3.0 - 4.9t^2 \]
03

Set up the equation for Ted

For Ted, we use the same equation but with different initial conditions: \[ x_T(t) = x_{T0} + v_{T0}t - \frac{1}{2}gt^2 \] where \(x_{T0} = 1.0 \) m, \(v_{T0} = 4.2 \) m/s, and \(g = 9.8 \) m/s². The equation becomes: \[ x_T(t) = 1.0 + 4.2t - 4.9t^2 \]
04

Final Equations

The final position-versus-time equations for Bill and Ted are:For Bill:\[ x_B(t) = 3.0 - 4.9t^2 \]For Ted:\[ x_T(t) = 1.0 + 4.2t - 4.9t^2 \]

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause this motion. In the context of the exercise, kinematics helps describe the motion of Bill and Ted as they move in space and time. The kinematic equations we use solve problems involving:
  • Displacement
  • Velocity
  • Acceleration
  • Time
These equations allow us to determine the future position of an object if we know certain initial conditions. For instance, the position-versus-time (\(x(t)\) equations) used in the solution for Bill and Ted help us understand their vertical motion after leaving the diving boards. Using the standard kinematic equation: \[x(t) = x_0 + v_0t + \frac{1}{2}at^2\]where \(x(t)\) is the position at time \(t\), \(x_0\) is the initial position, \(v_0\) is the initial velocity, and \(a\) is the acceleration. This formula is essential for predicting the trajectory of moving objects.
Initial Velocity
Initial velocity is the speed and direction an object starts moving at when it begins its motion. It plays a crucial role in determining how the object will move over time. In our exercise, Bill's initial velocity is 0 m/s because he simply steps off the diving board. Meanwhile, Ted has an initial velocity of 4.2 m/s as he jumps upward from his board.

This initial velocity is critical in calculating future positions using the kinematic equations. For Ted, his initial upward velocity means he'll ascend before the pull of gravity decelerates and then accelerates him downward. This contrasts with Bill’s situation where he immediately begins falling downward. These different initial conditions are applied in their respective equations:
  • \[x_B(t) = 3.0 - 4.9t^2\] for Bill
  • \[x_T(t) = 1.0 + 4.2t - 4.9t^2\] for Ted
Remember, understanding initial velocity helps predict how quickly an object will reach a certain point and its trajectory path.
Acceleration Due to Gravity
Acceleration due to gravity is a constant force that pulls objects toward the Earth's center. On Earth, its value is approximately 9.8 m/s². This downward acceleration is the same for both Bill and Ted, but it affects their motions differently due to their different starting velocities.

In the step-by-step solution, the effect of gravity is captured in the term \(-\frac{1}{2}gt^2\) of the kinematic equations, where \(g = 9.8\) m/s². This term describes how gravity causes a quadratic increase in velocity as time passes, influencing the downward motion of both Bill and Ted:
  • For Bill, since his initial velocity is zero, gravity is the only force changing his motion (\(-4.9t^2\)).
  • For Ted, gravity counters his initial upward motion, causing him to eventually fall back towards the water (\(4.2t - 4.9t^2\)).
Understanding gravity’s role is key to predicting and calculating the motions of objects falling or moving under its influence.

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Most popular questions from this chapter

Suppose the balloon is descending with a constant speed of \(4.2 \mathrm{m} / \mathrm{s}\) when the bag of sand comes loose at a height of \(35 \mathrm{m}\). (a) How long is the bag in the air? (b) What is the speed of the bag when it is \(15 \mathrm{m}\) above the ground?

It was a dark and stormy night, when suddenly you saw a flash of lightning. Three-and-a-half seconds later you heard the thunder. Given that the speed of sound in air is about \(340 \mathrm{m} / \mathrm{s}\) how far away was the lightning bolt?

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The position of a particle as a function of time is given by \(x=(-2.00 \mathrm{m} / \mathrm{s}) t+\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) (a) Plot \(x\) versus \(t\) for time from \(t=0\) to \(t=1,00 \mathrm{s}\) (b) Find the average velocity of the particle from \(t=0.150 \mathrm{s}\) to \(t=0.250 \mathrm{s},\) (c) Find the average velocity from \(t=0.190 \mathrm{s}\) to \(t=0.210 \mathrm{s}\) (d) Do you expect the instantaneous velocity at \(t=0.200 \mathrm{s}\) to be closer to \(-1.62 \mathrm{m} / \mathrm{s},\) or \(-1.66 \mathrm{m} / \mathrm{s}\) ? Explain.

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