91Ó°ÊÓ

A golf ball is hit off a tee at the edge of a cliff. Its \(x\) and \(y\) coordinates as functions of time are given by the following expressions: $$\begin{array}{c}x=(18.0 \mathrm{m} / \mathrm{s}) t \\ \text { and } \quad y=(4.00 \mathrm{m} / \mathrm{s}) t-\left(4.90 \mathrm{m} / \mathrm{s}^{2}\right) t^{2} \end{array}$$ (a) Write a vector expression for the ball's position as a function of time, using the unit vectors \(\mathbf{i}\) and \(\mathbf{j} .\) By taking derivatives, obtain expressions for (b) the velocity vector \(\mathbf{v}\) as a function of time and (c) the acceleration vector a as a function of time. Next use unit- vector notation to write expressions for (d) the position, (e) the velocity, and (f) the acceleration of the golf ball, all at \(t=3.00 \mathrm{s}\)

Step by step solution

01

Express the position vector

Firstly, we can express the ball's position \(r(t)\) as a function of time using the given equations. The x-component of the position is given by \(x = (18.0 m/s) t\) and the y-component is given by \(y = (4.00 m/s) t - (4.90 m/s^2) t^2\). Thus the position vector is given by \(r(t) = (18.0 m/s) t \mathbf{i} + \[(4.00 m/s) t - (4.90 m/s^2) t^2\] \mathbf{j}\)

02

Find the velocity vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector with respect to time. Differentiating \(r(t)\), we get: \(v(t) = \[\frac{dr(t)}{dt}\] = 18.0 m/s \mathbf{i} + \[(4.00 m/s - 2*(4.90 m/s^2) t)\] \mathbf{j}\)

03

Find the acceleration vector

The acceleration vector is the derivative of the velocity vector with respect to time. Differentiating \(v(t)\), we get: \(a(t) = \[\frac{dv(t)}{dt}\] = 0 \mathbf{i} + (-9.80 m/s^2)\mathbf{j} \). Here, the \( \mathbf{i} \)-component is 0 because the ball is not accelerating horizontally and the \( \mathbf{j} \)-component represents the acceleration due to gravity.

04

Find the position at \( t=3.00s \)

Substitute \(t=3s\) into the equation for \(r(t)\) from Step 1: \(r(3) = (18.0 m/s * 3s) \mathbf{i} + \[(4.00 m/s * 3s - 4.90 m/s^2 * (3s)^2)\] \mathbf{j} .\)

05

Find the velocity at \( t=3.00s \)

Substitute \( t=3s \) into the equation for \(v(t)\) from Step 2: \(v(3) = 18.0 m/s \mathbf{i} + \[(4.00 m/s - 2*(4.90 m/s^2) * 3s)\] \mathbf{j} .\)

06

The acceleration at \( t=3.00s \)

The acceleration vector does not depend on time, so it is constant and remains \(-9.80 m/s^2 \mathbf{j}\) at any time.

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Key Concepts

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

Position Vector

The position vector helps us determine where an object is located at any given time.
For a golf ball hit from a cliff, considering its horizontal and vertical displacement is crucial.
The position vector combines these two components.
The horizontal position is given by the x-component:

• Formula: \(x = (18.0 \text{ m/s})t\)
• Meaning: The ball travels 18 meters per second along the x-axis.

The vertical position is governed by the y-component:

• Formula: \(y = (4.00 \text{ m/s})t - (4.90 \text{ m/s}^2)t^2\)
• Explanation: The first term \(4.00 \text{ m/s}t\) shows its initial upward motion, while the second term \(4.90 \text{ m/s}^2 t^2\) represents the downward pull of gravity.

Combining these, we form the full position vector \(r(t) = (18.0\text{ m/s})t \mathbf{i} + [(4.00\text{ m/s})t - (4.90 \text{ m/s}^2)t^2] \mathbf{j}\) which tells us the ball's position in both directions at any time \(t\).

Velocity Vector

The velocity vector describes the speed and direction of an object’s motion at any given moment.
It is the derivative of the position vector with respect to time.
For the golf ball, differentiate the position vector components separately:

• The horizontal velocity is a constant: \(18.0 \text{ m/s}\) in the \(\mathbf{i}\) direction.
• The vertical velocity involves change due to gravity: \( v_y(t) = \frac{d}{dt}[(4.00\text{ m/s})t - (4.90 \text{ m/s}^2)t^2] = (4.00 \text{ m/s}) - (9.80 \text{ m/s}^2)t\)

Thus, the velocity vector is \(\mathbf{v}(t) = 18.0 \text{ m/s} \mathbf{i} + [(4.00 \text{ m/s}) - (9.80 \text{ m/s}^2)t] \mathbf{j}\)
This tells us how the ball's position changes from one point to the next as time progresses.

Acceleration Vector

Acceleration indicates how the velocity of an object changes over time.
It is the derivative of the velocity vector.
For the golf ball, we focus on how the velocity changes:

• The horizontal component remains zero: the ball’s horizontal velocity does not change.
• The vertical acceleration is due to gravity: \( a_y(t) = \frac{d}{dt}[(4.00\text{ m/s}) - (9.80 \text{ m/s}^2)t] = -9.80 \text{ m/s}^2\)

This yields a constant acceleration vector: \(\mathbf{a}(t) = 0 \mathbf{i} - 9.80 \text{ m/s}^2 \mathbf{j}\).
In essence, this vector always points downward, signifying the unchanging force of gravity acting on the ball.