Having a hard time remembering what this equation is

[LT72884]

Ok, so it has been about 4 years since i have had physics for scientists and engineers. I am a full time mechanical and areospace student. I just started dynamics class and right off the bat, i ran into an equation that i am not sure how to infer. It is from our rectilinear example:

d^2x/dt^2 = a

ok, so i know "a" is acceleration but on the LHS, i can't tell if it is telling me it is a second deriv with repsect to x or if it is distance squared multiplied by x? there is no explanation in the text.

Thanks:)

 

[magoo]

It should be the second derivative of x with respect to time.

 

[LT72884]

Ok, that helps a lot. Then the bottom is the deriv with respect to time squared?? man i forget all this stuff after 4 years

sometimes fonts in books or online, i can't tell what they are trying to tell me haha. ok. would you guys be willing to help me with an EXAMPLE problem. Bare in mind i have never had dynamics and not entirely sure what to think of it yet:)

I will list the example.
During a test, a rocket is traveling at 75m/s. When it reaches 40m, the engine fails. determine the max height Sb and the final speed of the rocket just before it hits the ground. Assume constant downward a and no air resistance.

I have no idea where to go with this problem. i don't even know where to start..well maybe i do. it has motion so it must involve one fo the BIG 3? but that's all i would know where to begin.

thanks a ton guys

 

[berkeman]

Ok, that helps a lot. Then the bottom is the deriv with respect to time squared?? man i forget all this stuff after 4 years

sometimes fonts in books or online, i can't tell what they are trying to tell me haha. ok. would you guys be willing to help me with an EXAMPLE problem. Bare in mind i have never had dynamics and not entirely sure what to think of it yet:)

I will list the example.
During a test, a rocket is traveling at 75m/s. When it reaches 40m, the engine fails. determine the max height Sb and the final speed of the rocket just before it hits the ground. Assume constant downward a and no air resistance.

I have no idea where to go with this problem. i don't even know where to start..well maybe i do. it has motion so it must involve one fo the BIG 3? but that's all i would know where to begin.

thanks a ton guys

To get help with schoolwork questions, you need to post in the Homework Help forums, and fill out the Template that you are provided there when you start a new thread. That Template includes sections for the Relevant Equations, and for your Attempt at a Solution. Just saying "I have no idea" is not allowed here.

Having said that, it still doesn't sound like you understand these basics:
v(t) = \frac{dx(t)}{dt}
a(t) = \frac{d^2x(t)}{dt^2}
The first equation says that the velocity of a particle is equal to the first derivative of the position as a function of time.

The second equation says that the acceleration of a particle is equal to the second derivative of the position as a function of time. There are no terms "squared" in the equation. The "2" that you see in the numerator and denominator of that equation mean to use the second derivative. (That means to differentiate twice in a row).

This thread is closed. Please do a bit more studying, and post your example problem in the Homework Help forums. Thanks.

 

[Mark44]

d^2x/dt^2 = a

ok, so i know "a" is acceleration but on the LHS, i can't tell if it is telling me it is a second deriv with repsect to x or if it is distance squared multiplied by x?

$$a(t) = \frac{d^2x(t)}{dt^2}$$The second equation says that the acceleration of a particle is equal to the second derivative of the position as a function of time.

To expand on what @berkeman said, this -- $\frac{d^2x(t)}{dt^2}$ -- is notation that means $\frac d{dt}\left(\frac{dx(t)}{dt}\right)$. It is a notational convenience more than anything else that we write this second derivative as $\frac{d^2x(t)}{dt^2}$.