Calculus question

[calculus19]

Hey all,

I was wondering if anyone here can help me out with a small problem I'm having regarding this question!

from this equation 8e^-t/3(-1/3t + 1) I'm supposed to find when the acceleration of the body first reaches zero. Can anyone help me out with this? I believe I have to solve for t? However I'm not sure what steps to take!

Any help will be much appreciated!

Thanks

 

[Quaid]

help me out with a small problem I'm having

from this equation 8e^-t/3(-1/3t + 1) I'm supposed to find when the acceleration of the body first reaches zero.

Is this question part of an assignment? Can you explain what the small problem is? (By the way, what is the question statement, including instructions?)

The expression 8e^(-t/3)(-1/3*t + 1) is not an equation, but, if you've been told that acceleration of the body at time t is defined by that expression, then, yes, you set it equal to zero and solve for t.

If you've not been told what acceleration is, from where did that expression come?

I've made assumptions, about the order of operations, shown in red.

8e^(-t/3)(-1/3*t + 1) = 0

If this is the correct equation, for the endeavor, then note that this equation shows that some product of two numbers is zero.

The first number is 8*e^(-t/3) and the second number is (-1/3*t + 1).

If you're familiar with the exponential function e^x, then you know that 8*e^(-t/3) is always a positive number (i.e., never zero).

So, look what we have in that equation: a positive number 8e^(-t/3) multiplied by the number (-1/3*t+1) equals zero.

Use the Zero-Product Property, to solve for t (remember the property -- if two numbers multiply to make zero, then at least one of them is zero).

Cheers

 

[calculus19]

Is this question part of an assignment? Can you explain what the small problem is? (By the way, what is the question statement, including instructions?)

The expression 8e^(-t/3)(-1/3*t + 1) is not an equation, but, if you've been told that acceleration of the body at time t is defined by that expression, then, yes, you set it equal to zero and solve for t.

If you've not been told what acceleration is, from where did that expression come?

I've made assumptions, about the order of operations, shown in red.

8e^(-t/3)(-1/3*t + 1) = 0

If this is the correct equation, for the endeavor, then note that this equation shows that some product of two numbers is zero.

The first number is 8*e^(-t/3) and the second number is (-1/3*t + 1).

If you're familiar with the exponential function e^x, then you know that 8*e^(-t/3) is always a positive number (i.e., never zero).

So, look what we have in that equation: a positive number 8e^(-t/3) multiplied by the number (-1/3*t+1) equals zero.

Use the Zero-Product Property, to solve for t (remember the property -- if two numbers multiply to make zero, then at least one of them is zero).

Cheers

Hi Quaid,

Thank you for taking the time to respond and clearing this issue up for me.

That expression came from an equation which I approached and worked through by using the product rule.

As you have shared some light into the correct action that I need to take, there's no need elaborate further on the question.

Many thanks!