Find Acceleration using Velocity and a given equation

In summary: But derivatives are really important to know and you will use them a lot in physics and other math courses. So definitely learn them as soon as you can. Good luck!In summary, the conversation revolved around solving a physics problem involving a race car's acceleration. The solution involved using derivatives, which the asker had not yet learned but planned on learning. The conversation also clarified that integrals are not necessary to learn at this point.
  • #1
destroctokon
2
0

Homework Statement



A race car starts from rest and travels east along a straight and level track. For the first 5.0s of the car's motion, the eastward component of the car's velocity is given by vx (t) = (0.920m/s^3) t^2.

What is the acceleration of the car when vx = 13.9m/s ?

Homework Equations


a= Δv/Δt


The Attempt at a Solution


I attempted to solve the problem by using the given formula for vx(t)
13.9 = (.92m/s^3)* t^2

With this i solved for t and got 3.89s. so using this I put it into the formula for a

(13.9 - .92)/(3.89-1) = 4.5m/s

Unfortunantly this was wrong. can anyone help me? Should I have subtracted 0 for both initial velocity and initial time? or was that general thought process right?

Thank you.
 
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  • #2
destroctokon said:

Homework Statement



A race car starts from rest and travels east along a straight and level track. For the first 5.0s of the car's motion, the eastward component of the car's velocity is given by vx (t) = (0.920m/s^3) t^2.

What is the acceleration of the car when vx = 13.9m/s ?

Homework Equations


a= Δv/Δt

The Attempt at a Solution


I attempted to solve the problem by using the given formula for vx(t)
13.9 = (.92m/s^3)* t^2

With this i solved for t and got 3.89s. so using this I put it into the formula for a

(13.9 - .92)/(3.89-1) = 4.5m/s

Unfortunantly this was wrong. can anyone help me? Should I have subtracted 0 for both initial velocity and initial time? or was that general thought process right?

Thank you.

I am unsure as to if you have learned how to do derivatives yet. But that is the easiest way of solving this. The derivative of displacement is velocity, and the derivative of velocity is acceleration. Have you by any chance learned derivatives yet?

I will explain using the derivative.
First thing you need to do is find the time when the speed is 13.9 m/s.
Fill it into the equation and isolate t as you did. I also got 3.89 s.

Next is getting the acceleration equation. Take the velocity equation.
V(x)=0.920t^2
The corresponding acceleration equation for this is in the form
a(x)= (Exp)(base numer)(t)^(exponent - 1)
a(x)=1.840t

Now input the time into the equation and you should get your answer.
 
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  • #3
Thank you for the help. Using the derivative I was able to get it right. Unforunatnly I have no yet learned how to do them. My professor spoke of them being used here, and I was intending on learning them on my own.

So I'm assuming I should learn how to do derivatives and integral as soon as possible to be able to do most if not all of my next homework problems.

But thank you for the help!
 
  • #4
destroctokon said:
Thank you for the help. Using the derivative I was able to get it right. Unforunatnly I have no yet learned how to do them. My professor spoke of them being used here, and I was intending on learning them on my own.

So I'm assuming I should learn how to do derivatives and integral as soon as possible to be able to do most if not all of my next homework problems.

But thank you for the help!

No problem. Happy to help. Just 1 thing though, you learn integrals in university or college, so you don't need to bother with learning them now. You could if you want, but it would be a lot of extra work.
 
  • #5




Your thought process was correct, but there are a few errors in your calculations. First, when you solved for t using the given formula, you should have taken the square root of both sides, as the equation is in the form of t^2. This would give you t = 3.68s.

Next, when finding the acceleration, you should use the final velocity (13.9m/s) and the initial velocity (0m/s), as the car starts from rest. So the formula would be (13.9m/s - 0m/s) / (3.68s - 0s) = 3.77m/s^2.

It's important to pay attention to units as well. In this case, the given velocity is in m/s^3, but the units for acceleration should be m/s^2. So when solving for t, you should have taken the cube root of both sides, and when finding the acceleration, you should have divided by the square of the time (3.68s^2).

Overall, your general thought process was correct, but pay attention to units and make sure to double check your calculations.
 

FAQ: Find Acceleration using Velocity and a given equation

What is acceleration?

Acceleration is the rate of change of velocity over time. It is a vector quantity, meaning it has both magnitude and direction. In other words, it is the amount by which an object's velocity changes in a certain amount of time.

2. How is acceleration related to velocity?

Acceleration is directly proportional to velocity. This means that as an object's velocity increases, its acceleration also increases. However, the direction of the acceleration may be different from the direction of the velocity.

3. What is the equation for finding acceleration using velocity?

The equation for finding acceleration using velocity is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. This equation is known as the average acceleration equation.

4. How do you use a given equation to find acceleration?

To use a given equation to find acceleration, you must first identify which variables are given and which variable you are trying to solve for. Then, plug in the given values into the equation and solve for the unknown variable. Make sure to pay attention to units and use the correct units for each variable.

5. Can acceleration be negative?

Yes, acceleration can be negative. This means that the object is slowing down, or decelerating, in the direction opposite to its initial motion. A negative acceleration can also be referred to as deceleration or retardation.

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