What is the Time When the Particle Has Zero Acceleration?

In summary, the acceleration of a particle moving along the x-axis can be calculated using the formula a(t) = 5[(10-7t^2)(-2)(10+7t^2)^(-3) *(14t) + (10+7t^2)^2 *(10-7t^2)] where t is the time in seconds and x is the position in meters. By factoring out common terms, the equation can be simplified to 5(10-7t^2) * (10+7t^2)^-2, and the acceleration can be found by finding the zeroes of the resulting equation.
  • #1
Destrio
212
0
A particle moves along the x-axis, its position at time t is given by
x(t)= 5t/(10+7t^2)

where t is >= 0 and measured in seconds and x is in meters

The acceleration of the particle equals 0 at time t = ?

I took the 1st derivative and got
i got v(t ) = 5(10-7t^2) / (10+7t^2)^2

or i got v(t ) = 5(10-7t^2) * (10+7t^2)^-2

then i took the derivative of that and got:

a(t ) = 5[(10-7t^2)(-2)(10+7t^2)^(-3) *(14t) + (10+7t^2)^2 *(10-7t^2)]

so i tried factoring out (10-7t^2), to get one answer which i got as sqrt(10/7)

and then for the other answer i got:
(10+7t^2)^5 = 28t

which i simplified to:
2401t^8 + 13720t^6 + 29400t^4 + 28000t^2 - 28t + 10000 = 0

now I'm not sure if I'm on the right track, and if I am, I'm not sure how I can solve for t.

Any help is much appreciated, I've been working on this (my first problem) for about an hour and I have a midterm in a few days :s

thanks
 
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  • #2
Destrio said:
v(t ) = 5(10-7t^2) * (10+7t^2)^-2

then i took the derivative of that and got:

a(t ) = 5[(10-7t^2)(-2)(10+7t^2)^(-3) *(14t) + (10+7t^2)^2 *(10-7t^2)]

Just keep the powers of (10+7t^2)^2 in the denominator like you did when calculating v. The numerator is then easy to factorize.
 
  • #3
the numerator will still end up being 2401t^8 + 13720t^6 + 29400t^4 + 28000t^2 - 28t + 10000
if i make a common denomintor with the two terms
ahhh
 
  • #4
Don't write out the brackets, that's going to be horrible. If you never write out the brackets in your calculations you'll end up with a form that's already quite nicely factorized.

If you apply to quotient rule to:
5(10-7t^2)/(10+7t^2)^2
without expanding, you get (10+7t^2)^4 in the denominator right? YOu'r numerator will not be too bad either. Just look for common terms.
 
  • #5
what do you mean don't write out the brackets?

applying the quotient rule i got:
5[(10+7t^2)^2 *(-14t) - (10-7t^2)(2)(10+7t^2)(14t)]/(10+7t^2)^4

so I can factor out 14t
and 10+7t^2

will my two terms be
10+7t^2
and
10-7t^2
?

thanks
 
  • #6
why did I get something different from when I used to product rule?
or were they equivalent but arranged differently?
 
  • #7
You made a small error when using the product rule. The last answer you gave using the qoutient rule is correct and it should be easy finding the zero's now.
 

What is the second derivative in calculus?

The second derivative in calculus is the derivative of the first derivative. In other words, it is the rate at which the slope of a function is changing. It measures the curvature of a function at a specific point.

Why is the second derivative important?

The second derivative is important because it helps us to better understand the behavior of a function. It can tell us whether a function is increasing or decreasing, concave up or down, and whether it has any points of inflection. It also helps us to find maximum and minimum values of a function.

How do you find the second derivative of a function?

To find the second derivative of a function, you first find the first derivative using the power rule, product rule, quotient rule, or chain rule. Then, you take the derivative of the first derivative using the same rules. This will give you the second derivative of the function.

What is the relationship between the second derivative and the graph of a function?

The second derivative can tell us a lot about the shape of a function's graph. If the second derivative is positive, the function is concave up, meaning the graph is shaped like a bowl. If the second derivative is negative, the function is concave down, meaning the graph is shaped like a hill. The second derivative can also help us identify points of inflection on a graph.

How can the second derivative be used in real-life applications?

The second derivative is used in many real-life applications such as physics, engineering, economics, and biology. It can help us to analyze rates of change, optimize functions, and make predictions about future behavior. For example, in economics, the second derivative can be used to determine the marginal cost and marginal revenue of a business. In physics, it can be used to calculate the acceleration of an object.

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