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Chapter 3: Problem 25

An object moves in the \(x y\) -plane. The \(x\) - and \(y\) -coordinates of the object as a function of time are given by the following equations: \(x(t)=4.9 t^{2}+2 t+1\) and \(y(t)=3 t+2 .\) What is the velocity vector of the object as a function of time? What is its acceleration vector at a time \(t=2\) s?

Short Answer

Expert verified
Answer: The velocity vector as a function of time is v(t) = [9.8t + 2, 3], and the acceleration vector at t=2s is [9.8, 0].

Step by step solution

01

Find the velocity vector as a function of time

To find the velocity vector, we need to take the derivative of the position functions with respect to time, t. The x(t) function is given by: x(t) = 4.9t^2 + 2t + 1 Taking the derivative with respect to time, we get the x-component of the velocity vector: v_x(t) = \frac{dx}{dt} = 9.8t + 2 The y(t) function is given by: y(t) = 3t + 2 Taking the derivative with respect to time, we get the y-component of the velocity vector: v_y(t) = \frac{dy}{dt} = 3 Now, we can combine the x and y components to form the velocity vector: v(t) = [9.8t + 2, 3]
02

Find the acceleration vector at t=2s

To find the acceleration vector, we need to take the derivative of the velocity functions with respect to time, t, or take the second derivative of the position functions. Taking the derivative of v_x(t) (which is the second derivative of x(t)) with respect to time, we get the x-component of the acceleration vector: a_x(t) = \frac{d^2x}{dt^2} = 9.8 Taking the derivative of v_y(t) (which is the second derivative of y(t)) with respect to time, we get the y-component of the acceleration vector: a_y(t) = \frac{d^2y}{dt^2} = 0 Now, we can combine the x and y components to form the acceleration vector at any given time t: a(t) = [9.8, 0] To find the acceleration vector at t=2s, we simply plug t=2 into the acceleration vector equation: a(2) = [9.8, 0] The acceleration vector at time t=2s is [9.8, 0].

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Key Concepts

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

Understanding Kinematics Equations
Kinematics equations are the backbone of classical mechanics, especially when it's about motion in physics. They link the position, velocity, and acceleration of an object with time, which are essential to understanding its motion.

Kinematics is not concerned with the forces that cause the motion but rather describes the motion itself. A primary set of equations that describe this motion in one dimension are:
  • The position equation: \( x(t) = x_0 + v_0t + \frac{1}{2}at^2 \)
  • The velocity equation: \( v(t) = v_0 + at \)
  • The acceleration equation: \( a(t) = a \), where \( a \) is constant.
But what if acceleration isn't constant? This leads us to our next key topic.
Time-Dependent Acceleration Concepts
While the kinematic equations assume constant acceleration, many real-world scenarios involve time-dependent acceleration, which means the acceleration changes as time progresses. To tackle this, calculus becomes an indispensable tool.

For time-dependent acceleration, you need to integrate acceleration over time to get the velocity, and similarly, integrate the velocity to find position. If you have an acceleration as a function of time, \( a(t) \), the velocity at time \( t \) is found by integrating \( a(t) \):
  • \( v(t) = \int a(t) \, dt \)
And for position:
  • \( x(t) = \int v(t) \, dt \)
The beauty of calculus lies in its ability to deal with such dynamic variables effectively. This concept is pivotal when acceleration is not uniform, which is the case in many complex motions.
Derivative of Position - Velocity Vector
In the realm of physics, particularly in the study of kinematics, the derivative of position with respect to time gives us an object's velocity vector. It is a vector because velocity has both magnitude and direction, which tells us not just how fast the object is traveling but also the direction of its travel.

Considering an object moving in the xy-plane, with its position described by functions \( x(t) \) and \( y(t) \), the velocity is found by differentiating these position functions with respect to time:
  • \( v_x(t) = \frac{dx}{dt} \)
  • \( v_y(t) = \frac{dy}{dt} \)
These components \( v_x(t) \) and \( v_y(t) \) form the velocity vector \( v(t) = [v_x(t), v_y(t)] \). Thus, if an object's position equations in terms of time are known, its velocity at any moment can be accurately determined by taking the time derivative of its position.

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Most popular questions from this chapter

An air-hockey puck has a model rocket rigidly attached to it. The puck is pushed from one corner along the long side of the \(2.00-\mathrm{m}\) long air- hockey table, with the rocket pointing along the short side of the table, and at the same time the rocket is fired. If the rocket thrust imparts an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) to the puck, and the table is \(1.00 \mathrm{~m}\) wide, with what minimum initial velocity should the puck be pushed to make it to the opposite short side of the table without bouncing off either long side of the table? Draw the trajectory of the puck for three initial velocities: \(vv_{\min } .\) Neglect friction and air resistance.

10.0 seconds after being fired, a cannonball strikes a point \(500 . \mathrm{m}\) horizontally from and \(100 . \mathrm{m}\) vertically above the point of launch. a) With what initial velocity was the cannonball launched? b) What maximum height was attained by the ball? c) What is the magnitude and direction of the ball's velocity just before it strikes the given point?

Two swimmers with a soft spot for physics engage in a peculiar race that models a famous optics experiment: the Michelson-Morley experiment. The race takes place in a river \(50.0 \mathrm{~m}\) wide that is flowing at a steady rate of \(3.00 \mathrm{~m} / \mathrm{s} .\) Both swimmers start at the same point on one bank and swim at the same speed of \(5.00 \mathrm{~m} / \mathrm{s}\) with respect to the stream. One of the swimmers swims directly across the river to the closest point on the opposite bank and then turns around and swims back to the starting point. The other swimmer swims along the river bank, first upstream a distance exactly equal to the width of the river and then downstream back to the starting point. Who gets back to the starting point first?

You passed the salt and pepper shakers to your friend at the other end of a table of height \(0.85 \mathrm{~m}\) by sliding them across the table. They both missed your friend and slid off the table, with velocities of \(5 \mathrm{~m} / \mathrm{s}\) and \(2.5 \mathrm{~m} / \mathrm{s}\), respectively. a) Compare the times it takes the shakers to hit the floor. b) Compare the distance that each shaker travels from the edge of the table to the point it hits the floor.

A circus juggler performs an act with balls that he tosses with his right hand and catches with his left hand. Each ball is launched at an angle of \(75^{\circ}\) and reaches a maximum height of \(90 \mathrm{~cm}\) above the launching height. If it takes the juggler \(0.2 \mathrm{~s}\) to catch a ball with his left hand, pass it to his right hand and toss it back into the air, what is the maximum number of balls he can juggle?
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