How Does Acceleration Affect Velocity and Position in Kinematics?

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The discussion revolves around a kinematics problem involving acceleration, velocity, and position. The acceleration function is given as a(t) = 3 m/s³ · t, leading to an increasing acceleration in the positive direction. The user initially miscalculated the velocity and position functions after integrating the acceleration, resulting in incorrect conclusions about their sizes and directions. After receiving feedback, the user realized a mistake in calculating the position at t=4 seconds and corrected their approach. Ultimately, the user successfully solved the problem with assistance from others.
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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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