/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A cart is propelled over an \(x ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A cart is propelled over an \(x y\) plane with acceleration components \(a_{x}=4.0 \mathrm{~m} / \mathrm{s}^{2}\) and \(a_{y}=-2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Its initial velocity has components \(v_{0 x}=8.0 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=12 \mathrm{~m} / \mathrm{s} .\) In unit-vector notation, what is the velocity of the cart when it reaches its greatest \(y\) coordinate?

Short Answer

Expert verified
The velocity of the cart is \(32 \hat{i} + 0 \hat{j} \text{ m/s}\).

Step by step solution

01

Understand the Greatest Y Coordinate Condition

The greatest y-coordinate is reached when the cart's vertical velocity component, \( v_y \), becomes zero. This is because a positive vertical velocity indicates upward motion, and reaching zero vertical velocity means the cart has reached its highest point.
02

Use the Vertical Velocity Equation

The equation for velocity in the vertical direction is given by \( v_y = v_{0y} + a_y t \). Since we need \( v_y = 0 \) at the maximum height, we can set this equation to zero and solve for \( t \):\[0 = 12 \, \text{m/s} - 2 \, \text{m/s}^2 \cdot t\]
03

Solve for Time to Reach Maximum Y

Solve the equation from Step 2 for \( t \):\[0 = 12 - 2t \2t = 12 \t = 6 \, \text{seconds}\]This is the time it takes for the cart to reach its maximum y-coordinate.
04

Find the Horizontal Velocity Component

The horizontal velocity component \( v_x \) is unaffected by the vertical motion and can be calculated using:\[v_x = v_{0x} + a_x t\]Substitute the values:\[v_x = 8 \, \text{m/s} + 4 \, \text{m/s}^2 \times 6 \, \text{s} = 8 + 24 = 32 \, \text{m/s}\]
05

Vectorize the Velocity

The velocity of the cart at the greatest y-coordinate is only in the horizontal direction since \( v_y = 0 \). Therefore, the velocity vector in unit-vector notation is:\[\vec{v} = 32 \hat{i} + 0 \hat{j} \, \text{m/s}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Velocity
Understanding velocity is key to kinematics, the branch of physics focused on the motion of objects. Velocity is a vector quantity, which means it has both magnitude (speed) and direction. In the context of our problem, we are observing the cart's velocity in terms of its horizontal and vertical components.
  • Horizontal Velocity (\( v_x \)): This remains constant unless acted upon by an external force. It's inclined by the horizontal acceleration (\( a_x \)). In this case, the equation is:\[ v_x = v_{0x} + a_x t \]
  • Vertical Velocity (\( v_y \)): Affected by factors like gravity (or any vertical acceleration). For the cart, when \( v_y = 0 \), it means the cart is at its highest point.
The cart will have a velocity when reaching this peak, largely dominated by its horizontal component since \( v_y = 0 \). Thus, understanding velocity involves acknowledging how these components interact over time.
Acceleration
Acceleration indicates how quickly velocity changes with time. It's also a vector quantity, with both magnitude and direction, and plays a vital role in determining how the cart's motion evolves.
  • Horizontal Acceleration (\( a_x \)): It shows how the cart speeds up on the x-axis. In this problem, \( a_x = 4.0 \, \text{m/s}^2 \), meaning the cart is gaining speed horizontally.
  • Vertical Acceleration (\( a_y \)): Since \( a_y = -2.0 \, \text{m/s}^2 \), it's acting against the upwards initial velocity, causing the cart to decelerate vertically until it stops rising and starts descending.
Recognizing these acceleration components is crucial for analyzing how the cart moves and how to compute when it reaches its maximum height.
Projectile Motion
Projectile motion combines both horizontal and vertical motion, creating a trajectory. In this scenario, the cart experiences projectile motion as it's propelled over a plane.
- **Characteristics of Projectile Motion**: - It has both an initial upward velocity and a downward acceleration due to gravity or another force. - The horizontal motion is uniform, meaning there's no acceleration affecting it initially in our ideal case except external forces.The greatest y-coordinate or maximum height is determined by when the vertical component of velocity (\( v_y \)) equals zero. Once this peak is reached, the vertical velocity reverses direction, and the cart begins to descend, continuing the characteristic path of projectile motion.
Unit-Vector Notation
Unit-vector notation is a method of expressing vectors, fundamental for understanding and simplifying calculations involving vector quantities like velocity and acceleration.
- **Components of Unit-Vector Notation**: - The horizontal component is described using \( \hat{i} \). - The vertical component is represented by \( \hat{j} \).In unit-vector notation, a vector is written as a sum of its horizontal and vertical components. For instance, the velocity vector at its highest point in our problem is expressed as:\[ \vec{v} = 32 \hat{i} + 0 \hat{j} \, \text{m/s} \] This notation is valuable for solving problems, as it clearly conveys the influence of each directional component. It's especially useful for interpreting motion in two dimensions, like the cart's path in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(4.0 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0 \mathrm{~m} / \mathrm{s},\) what is the smallest angle it can have for the pass to be legal?

A baseball leaves a pitcher's hand horizontally at a speed of \(161 \mathrm{~km} / \mathrm{h} .\) The distance to the batter is \(18.3 \mathrm{~m} .\) (a) How long does the ball take to travel the first half of that distance? (b) The second half? (c) How far does the ball fall freely during the first half? (d) During the second half? (e) Why aren't the quantities in (c) and (d) equal?

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of \(v_{0}=28.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(\theta_{0}=40.0^{\circ} .\) What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?

A car travels around a flat circle on the ground, at a constant speed of \(12.0 \mathrm{~m} / \mathrm{s}\). At a certain instant the car has an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) toward the east. What are its distance and direction from the center of the circle at that instant if it is traveling (a) clockwise around the circle and (b) counterclockwise around the circle?

Two ships, \(A\) and \(B\), leave port at the same time. Ship \(A\) travels northwest at 24 knots, and ship \(B\) travels at 28 knots in a direction \(40^{\circ}\) west of south. \((1 \mathrm{knot}=1\) nautical mile per hour; see Appendix D.) What are the (a) magnitude and (b) direction of the velocity of ship \(A\) relative to \(B ?\) (c) After what time will the ships be 160 nautical miles apart? (d) What will be the bearing of \(B\) (the direction of \(B\) 's position) relative to \(A\) at that time?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.