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 3, Problem 26P
To determine
The direction of average acceleration.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Question 1-16
A ball is launched horizontally with a velocity of 4.0 m/s from a height of 1.2 m above the ground. Assume air resistance is negligible. The time it takes for the ball to reach the ground is 0.50 s.
v = 4.0/s
V
1.2 m
When the ball strikes the ground, it will have a horizontal displacement (range) of -
0.50 meters.
1.2 meters.
2.0 meters.
8.0 meters.
A person jumps off a diving board 4.6 m above the water's surface into a deep pool. The person's downward motion stops 1.6 m below the surface of the water. Estimate the average deceleration (magnitude) of the person while under the water.
Use the definition to find an expression for the instantaneous velocity of an object moving with rectilinear motion according to the given functions relating s and t.
s=³ - 81?
The expression for the instantaneous velocity of the object is
Chapter 3 Solutions
Physics for Scientists and Engineers
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 60PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 68PCh. 3 - Prob. 69PCh. 3 - Prob. 70PCh. 3 - Prob. 71PCh. 3 - Prob. 72PCh. 3 - Prob. 73PCh. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - Prob. 80PCh. 3 - Prob. 81PCh. 3 - Prob. 82PCh. 3 - Prob. 83PCh. 3 - Prob. 84PCh. 3 - Prob. 85PCh. 3 - Prob. 86PCh. 3 - Prob. 87PCh. 3 - Prob. 88PCh. 3 - Prob. 89PCh. 3 - Prob. 90PCh. 3 - Prob. 91PCh. 3 - Prob. 92PCh. 3 - Prob. 93PCh. 3 - Prob. 94PCh. 3 - Prob. 95PCh. 3 - Prob. 96PCh. 3 - Prob. 97PCh. 3 - Prob. 98PCh. 3 - Prob. 99PCh. 3 - Prob. 100PCh. 3 - Prob. 101PCh. 3 - Prob. 102PCh. 3 - Prob. 103PCh. 3 - Prob. 104PCh. 3 - Prob. 105PCh. 3 - Prob. 106PCh. 3 - Prob. 107PCh. 3 - Prob. 108PCh. 3 - Prob. 109PCh. 3 - Prob. 110PCh. 3 - Prob. 111PCh. 3 - Prob. 112PCh. 3 - Prob. 113PCh. 3 - Prob. 114PCh. 3 - Prob. 115PCh. 3 - Prob. 116PCh. 3 - Prob. 117PCh. 3 - Prob. 118PCh. 3 - Prob. 119PCh. 3 - Prob. 120PCh. 3 - Prob. 121PCh. 3 - 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
- The position of a particle moving along the x -axis varies with time according to x(t)=5.0t24.0t3 . Find (a) the velocity and acceleration of the particle as functions of time, (b) the velocity and acceleration at t=2.0s , (c) the time at which the position is a maximum, (d) the time at which the velocity Is zero, and (e) the maximum position.arrow_forwardThe velocity of a particle moving along a straight line is given by v = 25t²- 80t -200 where v is measured in meters per second and t in seconds. It is given that the object is located 100 m to the left of the origin at t = 0s. Compute velocity when acceleration is zero position(s) the object changes direction.arrow_forwardSolve the velocity and displacement.arrow_forward
- 2-18) A plane flies for 3 h and reaches a point 600 km north and 800 km east of its starting point. Find the direction and magnitude of its average velocity. 2-19 A plane flies south at 500 km h¹ for 2 h and then flies west at 500 km h west for I h. (a) What is its average speed? (b) What are the mag- nitude and direction of its average velocity?arrow_forwardA football player kicks a ball at an angle of 36 degrees from the horizontal with an initial speed of 15.5 m/s. Assuming that the ball moves in a vertical plane, finda) The time, t 1 at which the ball reaches the highest point of its trajectory.b) Its maximum heightc) Its time of flight and range, andd) Its velocity when it strikes the groundarrow_forwardHow do you calculate the total time a projectile is in flight given only the initial velocity and maximum height?arrow_forward
- The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = vosin(e) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: Vfinal-y = Vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is Vfinal-y Then, = Vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: t Thus, the time to reach the maximum height is tmax-height= We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-yt + (1/2)ayt2 substituting, the vinitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following hmax = ( ) + (1/2)ayl Substituting ay = -g, results to hmax =( )- (1/2)g( simplifying the expression, yields hmax = x sinarrow_forwardGiven the velocity of an object and its initial position, how do you find the position of the object, for t ≥ 0?arrow_forwardThe acceleration of Particle moving along horizontal path is constsnt and equal to ( 4 m/sec") to the left. During certain time interval the particle has displacement of (42m) to the right and travels a total distance of (58m). Determine the initial and final velocity of the particle and the time interval.arrow_forward
- Determine the particle’s position and velocity at t = 1.00 s.arrow_forwardThe position on of a particle along the y axis is given by the following equation. Is the any time interval during which the particle is not moving?arrow_forwardThe position of a particle in millimeters is given by s = 27 – 12t + t, where t is in seconds. Plot the s-t and v-t relationships for the first 9 seconds. Determine the net displacement As during that interval and the total distance D traveled. By inspection of the s-t relationship, what conclusion can you reach regarding 3. the acceleration?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
University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY