Integral for displacement from velocity

In summary, the conversation revolved around calculating the magnitude of displacement for an object moving with a given velocity equation. After discussing various attempts at solving the problem, it was determined that the correct magnitude of displacement is 5, with the mistake in previous attempts being a result of incorrect arithmetic.
  • #1
burton95
54
0

Homework Statement


An object is moving with a velocity by the equation

v(t) = [(3/2)(m/s2)t] i^ + [(3/2)(m/s3)t2] j^

What is the magnitude of displacement during 0 - 2s

Homework Equations



v(t) = [(3/2)(m/s2)t] i^ + [(3/2)(m/s3)t2] j^

The Attempt at a Solution



(3/2) ∫ from 0 to 2 [(t2 / 2) i^] + [t3/3]y^

plug and chug with t = 2

(3/2) √(squaring each of the i^ + j^)

I end up with some decimal answer which I know is wrong. Where am I screwing up?
 
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  • #2
Who says you're wrong. What do you get? Is it something like 6.5?
 
  • #3
When I plug in 6.5 to the online quiz it says sorry wrong answer, "Don't forget to add the components of a vector quadratically to determine it's magnitude." I don't understand where I'm going wrong. Help
 
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  • #4
wait what about the units of s^2 and s^3?

EDIT: I don't think that's it
 
Last edited:
  • #5
Oops. 6.5 was wrong. I must have made a mistake somewhere. Here are the correct results: What did you get for the two components of the displacement vector? I got 3 and 4 using your equation. They were 3/2 x 2, and 3/2 x 8/3. So the correct displacement magnitude must be 5.

Sorry for any confusion I caused. It looks like you had it right all the way, but just made a mistake in arithmetic.

Chet
 

FAQ: Integral for displacement from velocity

What is the definition of an integral for displacement from velocity?

The integral for displacement from velocity is a mathematical concept used to calculate the total change in position over a given time period based on an object's velocity. It involves finding the area under the velocity-time curve, where the x-axis represents time and the y-axis represents velocity.

How is the integral for displacement from velocity different from a normal integral?

The integral for displacement from velocity is a specific type of integral that is used to solve problems related to displacement and velocity. It differs from a normal integral in that it involves a specific set of variables and is used for a specific purpose, rather than being a general mathematical tool.

What are the units of the integral for displacement from velocity?

The units of the integral for displacement from velocity depend on the units of velocity and time used in the problem. Generally, if the velocity is given in meters per second and time is given in seconds, the units of the integral will be in meters. However, it is important to check the units in each individual problem to ensure accuracy.

How is the integral for displacement from velocity applied in real-world situations?

The integral for displacement from velocity is commonly used in physics and engineering to solve problems related to an object's motion. It can be used to calculate the distance traveled by a car or the displacement of an object undergoing constant acceleration. It is also used in calculus to find the total change in position over a given time period.

What are some common mistakes to avoid when using the integral for displacement from velocity?

Some common mistakes when using the integral for displacement from velocity include not understanding the meaning and purpose of the integral, using incorrect units, and not properly setting up the integral based on the given problem. It is important to have a strong understanding of the concept and to carefully check for errors when solving problems using the integral for displacement from velocity.

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