COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 2, Problem 62QAP
To determine
The equation to relate position, speed and acceleration without the time variable.
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Students have asked these similar questions
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.
The velocity of an object is given by,
= u + at
V
ds
dt.
where, u is the initial velocity, a is the
acceleration, and ₺ is the time in seconds.
Solve the differential equation to obtain the
displacement, s, given that at t = 0, s = 7
(metres)
:
1
if u = 3 ms ¹, a = 2.4 ms 2, calculate the
displacements at t = 4.9 seconds.
Give your answer to 2 decimal places.
A quadratic fit is done on a distance versus time graph of an object with constant acceleration using a computer program. The distance is in meters and the time in seconds. The
form of the quadratic fit is y (z)= Az + Bz + Cand the fit parameters were found to by A-0.5, B-2 and C-3. What is the acceleration?
1.5 m/s
0.25 m/s
0.5 m/s
-2 m/s
1 m/s
3 m/s
OO0O C
Chapter 2 Solutions
COLLEGE PHYSICS
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Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY