Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 2, Problem 103P
(a)
To determine
To Sketch: velocity-time graph.
To Find: The area under the curve for the given time interval.
(b)
To determine
To obtain: The position function
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Given that the acceleration vector is a(t)=(−16cos(−4t))i+(−16sin(−4t))j+(1t)k, the initial velocity is v(0)=i+k, and the initial position vector is r(0)=i+j+k, compute:
A. The velocity vector v(t)
B. The position vector r(t)
the coordinates of an object moving in an xy plane vary with the time according to the eqations x=-7.85 sin w(omega)t, and y=4-7.85 cosw(omega)t. W(omega) is constant , x,y - in meters. And t in seconds.
Write expressions for the possition,velocity and acceleration vectors of the object at any time t>0.
A point particle of mass m = 1.8 kg moves according to the position function: r(t) = xtai + ytbj + ztck, where t denotes time and x, y, z, a, b, and c are constants such that the exponents are positive integers and the position function has the dimension of length.
Part (a) We can write the particle’s velocity function in the form v(t) = ntdi + otej + ptgk. Enter an expression for n in terms of x, y, z, a, b, and c.
Part (b) The particle’s velocity function will have the form v(t) = ntdi + otej + ptgk. Enter an expression for d in terms of x, y, z, a, b, and c. Part (c) Here is a set of parameter values for the motion of the particle: m = 1.8 kg, x = 1.8 m/s0, y = 2.4 m/s1, z = 0.15 m/s2, a = 0, b = 1, c = 2. Calculate the x-component of the particle’s angular momentum, in units of kg˙m2/s, about the origin at time t = 1 s. Part (d) Use the same set of parameter values (m = 1.8 kg, x = 1.8 m/s0, y = 2.4 m/s1, z = 0.15 m/s2, a = 0, b = 1, c = 2) to calculate the y-component of…
Chapter 2 Solutions
Physics for Scientists and Engineers
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 29PCh. 2 - Prob. 30PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 66PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 92PCh. 2 - Prob. 93PCh. 2 - Prob. 94PCh. 2 - Prob. 95PCh. 2 - Prob. 96PCh. 2 - Prob. 97PCh. 2 - Prob. 98PCh. 2 - Prob. 99PCh. 2 - Prob. 100PCh. 2 - Prob. 101PCh. 2 - Prob. 102PCh. 2 - Prob. 103PCh. 2 - Prob. 104PCh. 2 - Prob. 105PCh. 2 - Prob. 106PCh. 2 - Prob. 107PCh. 2 - Prob. 108PCh. 2 - Prob. 109PCh. 2 - Prob. 110PCh. 2 - Prob. 111PCh. 2 - Prob. 112PCh. 2 - Prob. 113PCh. 2 - Prob. 114PCh. 2 - Prob. 115PCh. 2 - Prob. 116PCh. 2 - Prob. 117PCh. 2 - Prob. 118PCh. 2 - Prob. 119PCh. 2 - Prob. 120PCh. 2 - Prob. 121PCh. 2 - Prob. 122P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- An object has an initial velocity of 29.0 m/s at 95.0° and an acceleration of 1.90 m/s2 at 200.0°. Assume that all angles are measured with respect to the positive x-axis. (a) Write the initial velocity vector and the acceleration vector in unit vector notation. (b) If the object maintains this acceleration for 12.0 seconds, determine the average velocity vector over the time interval. Express your answer in your unit vector notation.arrow_forwardAt time t = 0 a particle at the origin of an xyz-coordinate system has a velocity vector of Vo =i+ 12j - k. The acceleration function of the particle is a(t) = 16ti+j+ (cos 2t)k. Find the speed of the particle at time t = 1. Round your answer to two decimal places. Speed =arrow_forwardI need to solve v(sub y)=v(sub 0)sin θ(sub 0)-gt to find the time that it takes the projectile to reach maximum height (so I believe I'm solving for t?). I have previously found v(sub 0y) in part (a), but I'm unsure what v(sub 0) and v(sub y) are. How do I find those variables?arrow_forward
- At time t = 0 a particle at the origin of an xyz-coordinate system has a velocity vector of vo = i+ 5j – k. The acceleration function of the particle is a(t) = 32r²i + j+ (cos 21)k. Find the speed of the particle at time t = 1. Round your answer to two decimal places.arrow_forwardUse the given acceleration function and initial conditions to find the velocity vector v(t), and position vector r(t). Then find the position at time t = 5. a(t) = 71 + 4k v(0) = 6j, r(0) = 0 v(t) = r(t) = r(5) = Need Help? Read Itarrow_forwardSuppose that a particle begins its motion from the origin and meets the velocity (in m / s) given by: vx (t) = 2.50t vy (t) = 3.50√t. for t> 0 in seconds. Find the speed of the particle at t = 5.00 s What is the acceleration vector (in rectangular components) at t = 2.00 s ? Find the magnitude of the displacement between t = 0.00 s and t = 6.00 sarrow_forward
- A novice golfer on the green takes three strokes to sink the ball. The successive dis- placements are 6.6 m to the north, 5 m 45◦ north of east, and 6.6 m 17◦ west of south. Starting at the same initial point, an expert (lucky) golfer could make the hole in a single displacement. What is the magnitude of this single dis- placement? Answer in units of m.arrow_forwardThe figure shows the path taken by a drunk skunk over level ground, from initial point i to final point f. The angles are 01-32.0°, e2 = 49.0°, and 03 = 84.0°, and the distances are di 4.80 m, d2 = 7.30 m, and d3 = 10.0 m. What are the (a) magnitude and (b) angle of the skunk's displacement from i to f? Give the angle as a positive (counterclockwise) or negative (clockwise) angle of magnitude less than 180°, measured from the +x direction. dz dg (a) Number 5.586 Units (b) Number 135 Units * (degrees)arrow_forwardThe position of a particle moving in the xy plane varies with time, and its coordinates are given by the following expressions: x(t) = 4.00 m +r cos[(4.00/s)t] and y(t) = r sin[(4.00/s)t], where r = 2.00 m, and x and y will be in meters when t is in seconds. Determine the following for this particle. (a) speed of the particle at any time m/s (b) magnitude of the acceleration of the particle at any time m/s?arrow_forward
- The range of the projectile motion R in meters is given by the formula vo2 sin 20 where vo m/s is the initial velocity, g m/s² is the R = acceleration due to gravity and 0 is the radian measure of the angle makes with the horizontal at which the projectile is launched. What is the value of 0 that makes the range maximum? 3 4 6. the height of the right circular cylinder of greatest volume that can be inscribed inside the right circular cone with radius 4 in and height 12 in.arrow_forwardAt time t = 0, the position vector of a particle moving in the x-y plane is r = 4.62i m. By time t = 0.011 s, its position vector has become (4.84i +0.54j) m. Determine the magnitude Vay of its average velocity during this interval and the angle 0 made by the average velocity with the positive x-axis. Answers: Vav 0= i i m/sarrow_forwardThe equation r(t) = ( sin t)i + ( cos t)j + (t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. π Then write the particle's velocity at t= as a product of its speed and direction. The velocity vector is v(t) = (i+j+ k.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY