Answered: A particle travels along a straight…

Name: A particle travels along a straight line with a constantacceleration.
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: A particle travels along a straight line with a constantacceleration. When s = 4 ft, v = 3 ft/s and when s = 10 ft,v = 8 ft/s. Determine the velocity at S=16* =V

Science

Physics

A particle travels along a straight line with a constant acceleration. When s = 4 ft, v= 3 ft/s and when s= 10 ft, v = 8 ft/s. Determine the velocity at S=16

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter2: Kinematics

Section: Chapter Questions

Problem 1PE: Find the following for path A in Figure 2.59: (a) The distance traveled. (b) The magnitude of the...

See similar textbooks

Similar questions

We are standing on the top of a 1040 feet tall building and launch a small object upward. The object's height, measured in feet, after t seconds is h(t) = 16t? + 128t + 1040. A) What is the object initial velocity? ft/second B) What is the highest point that the object reaches? feet
A particle moves with a constant speed of 3 m/sec along the path shown where the equation of the path is given by y = 3x². What is the acceleration a at posi- tion x = 1.5 m? Give the rectangular (i, j) components of a. y = 3x²
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32o ramp at a speed of 40.0 m/s . How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are each 35.0 m long? ~ 10 ~ 7 ~ 4 ~ 12
A student fires a cannonball vertically upwards with a speed of 31.0m/s. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the cannonball. What was the cannonball's maximum height? 48.98 How long did the cannonball rise? 3.16 What was the cannonball's total flight time? 6.32
In many experiments, a velocity versus position (v-s) profile is obtained. v (ft/s) v = 0.6 s3/4 75 If we have a v-s graph for the tank truck, how can we – s(ft) determine its acceleration at 625 1500 position s = 1500 feet?
The coordinates of a particle moving along a curve are x(t) = -2t2 +15 and y(t) = t2 -10t + 15 , m whent is in seconds. Calculate the magnitude of velocity and acceleration when t= 5 sec.
The velocity of a particle is given by v = 23t2 - 110t + 52, where v is in meters per second and t is in seconds. Plot the velocity v and acceleration a versus time for the first 6.4 seconds of motion and evaluate the velocity when a is zero. Make the plots and then answer the questions. Questions: When t = 0.8 s, V = i m/s, a = i m/s2 When t = 3.7 s, V = i m/s, a = i m/s? When t = 4.7 s, V = i m/s, a = i m/s? When a = 0, V = m/s
A particle moves along two dimensions based on the following position vector r= [2.0 m + (3.00 m/s)f ]i + [(3.0 m)t - (2.00 m/s2)rlj a. Find the distance it covered in the first minute. b. Find the general expression of the instantaneous acceleration of the particle.
The particle travels along the path defined by the parabola y = 0.5x², where and y are in ft. (Figure 1) Figure 1 of 1 -y = 0.5x² If the component of velocity along the x axis is v₂ = (8t) ft/s, where t is in seconds, determine the particle's distance from the origin O t = 3 s. When t=0, x=0, y = 0. Express your answer using three significant figures and include the appropriate units. OI μA ? d = Value Units Submit Request Answer Part B Determine the magnitude of its acceleration when t = 3 s. Express your answer using three significant figures and include the appropriate units. μA ? a = Value Units Submit Provide Feedback Next > Request Answer
Find the following for path A in Figure 2.59: (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
My question isn't how to solve the problem exactly. In fact, it's already been solved on this website. My question is about the acceleration. When I solve this problem myself, first I calculate the velocity by dividing 100m by 53s. I get 1.89m/s. Then I use that to find the acceleration using the equation vf = vi + at. That's 1.89/53 = 0.036m/s^2. That's not correct. The correct way to find the acceleration is to us the equation d = 1/2 at^2 and solve that way without taking the intermediate step of finding the velocity. Doing it that way, the acceleration is 0.0712m/s^2. My question is why you get a different result doing it the first way than you get doing it the second way.
A particle is moving with a velocity of 60m/s in the positive x-direction at t = 0s. Between t = 0s andt = 15s, the velocity decreases uniformly to zero. What was the acceleration during this 15s interval? Whatis the signi cance of the sign of your answer?