1D Kinematics question

In summary: It's important to pay attention to the exponents when taking derivatives and integrals. Keep practicing and you'll get the hang of it. Good luck!
  • #1
Spartan029
15
0
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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  • #2
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.
 
  • #3
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!
 
  • #4
Sure, glad to help!
 

1. What is 1D Kinematics?

1D Kinematics is a branch of physics that deals with the motion of objects along a straight line. It involves studying various parameters such as position, velocity, and acceleration to describe the motion of an object.

2. How is 1D Kinematics different from 2D or 3D Kinematics?

1D Kinematics only involves motion along a single straight line, while 2D and 3D Kinematics involve motion in two or three dimensions respectively. This means that 1D Kinematics only considers motion in one direction, while 2D and 3D Kinematics consider motion in multiple directions.

3. What are the key equations used in 1D Kinematics?

The key equations used in 1D Kinematics are the equations of motion, which include:
- Position: x = x0 + v0t + 1/2at2
- Velocity: v = v0 + at
- Acceleration: a = (v - v0)/t
- Displacement: Δx = vavgt
- Average velocity: vavg = (v0 + v)/2
- Average acceleration: aavg = Δv/Δt

4. How is 1D Kinematics used in real life?

1D Kinematics is used in a variety of real-life scenarios, such as predicting the trajectory of a projectile, analyzing the motion of a car on a straight road, or calculating the speed and acceleration of a rollercoaster. It is also used in sports to analyze the motion of athletes and in engineering to design efficient and safe transportation systems.

5. What are some common misconceptions about 1D Kinematics?

One common misconception about 1D Kinematics is that an object with a constant velocity must also have a constant acceleration. This is not true, as an object can have a constant velocity but still be accelerating if its direction of motion is changing. Another misconception is that objects with a higher mass will always have a greater acceleration, but this is not always the case as an object's acceleration also depends on the force acting on it.

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