How Does Acceleration Affect Velocity and Position in Kinematics?

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Homework Help Overview

The discussion revolves around a kinematics problem involving acceleration, velocity, and position of an object. The acceleration is defined by a function of time, and participants are tasked with analyzing its implications on velocity and position over time.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the acceleration function and its effects on velocity and position, expressing confusion over the results of their calculations. They question the correctness of their interpretations regarding the size and direction of both acceleration and velocity.

Discussion Status

Some participants have provided feedback on the calculations, specifically pointing out a potential error in the original poster's evaluation of the position at a certain time. There is an ongoing exploration of the relationships between acceleration, velocity, and position, with no explicit consensus reached yet.

Contextual Notes

The problem includes specific initial conditions and requires integration to find velocity and position functions. Participants are navigating through the implications of these conditions and the mathematical operations involved.

Spartan029
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Hello everyone. I am stumped here with this problem, i feel like it should be fairly simple but i can't seem to figure it out.

Homework Statement



If the acceleration for a given object is given by the function:

a(t) = +(3 m/s^3) · t
(Note: units are included in the eqn, so if [t]=s then [a]=m/s^2.)

(a) Examining this function, you can tell that:
a1)the size of the acceleration will: ?
a2)the direction of the acceleration will be: ?

(b) If vi (at t=0 s) is -3 m/s, find v(t). Then answer the below:
(b1)the size of the velocity will: ?
(b2)the direction of the velocity will be: ?

(c) If xi (at t=0 s) is -7 m, find x(t). Then answer the below:
c1)After 4 s have passed, What position (x) is the object at now? ___m
c2)What is the total distance traveled (since t=0 s)? ___m

Homework Equations



I really don't know, Integration? maybe


The Attempt at a Solution



Okay so for part (a) I answered that the size would be "increasing" which is correct. Also the direction of the acceleration will be "in the + direction" which is aslo correct.
So for part (b) i take the integral of the a(t) and get:

v(t) = (3/2)t^2 - 3 m/s

by examining this i answered that the size would be "increasing" and "in the + direction" both of my answers are wrong and i don't understand why.

Okay so i skip to part (c) take the integral of v(t):

x(t) = (1/2)t^3 - 3t - 7 m

so
x(4) = (1/2)(4^3) - 3(4) - 7 = -11

-11 was not correct. :biggrin: so i thought it wise not to go on to part c2) lol

and here i am stumped. Any help would be much appreciated. Thanks!
 
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Hi Spartan029,

Spartan029 said:
Hello everyone. I am stumped here with this problem, i feel like it should be fairly simple but i can't seem to figure it out.

Homework Statement



If the acceleration for a given object is given by the function:

a(t) = +(3 m/s^3) · t
(Note: units are included in the eqn, so if [t]=s then [a]=m/s^2.)

(a) Examining this function, you can tell that:
a1)the size of the acceleration will: ?
a2)the direction of the acceleration will be: ?

(b) If vi (at t=0 s) is -3 m/s, find v(t). Then answer the below:
(b1)the size of the velocity will: ?
(b2)the direction of the velocity will be: ?

(c) If xi (at t=0 s) is -7 m, find x(t). Then answer the below:
c1)After 4 s have passed, What position (x) is the object at now? ___m
c2)What is the total distance traveled (since t=0 s)? ___m

Homework Equations



I really don't know, Integration? maybe


The Attempt at a Solution



Okay so for part (a) I answered that the size would be "increasing" which is correct. Also the direction of the acceleration will be "in the + direction" which is aslo correct.
So for part (b) i take the integral of the a(t) and get:

v(t) = (3/2)t^2 - 3 m/s

by examining this i answered that the size would be "increasing" and "in the + direction" both of my answers are wrong and i don't understand why.

Okay so i skip to part (c) take the integral of v(t):

x(t) = (1/2)t^3 - 3t - 7 m

so
x(4) = (1/2)(4^3) - 3(4) - 7 = -11

I think you squared the 4 instead of cubing it.
 
alphysicist said:
Hi Spartan029,



I think you squared the 4 instead of cubing it.


lol, yea i messed up there, thanks for pointing it out.

I figured out how to solve the problem thanks for helping me!
 
Sure, glad to help!
 

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