A weird velocity/acceleration question

[inaiki]

Homework Statement

This problem showed up in my final review packet, and I /think/ it should be basic kinematics, but I don't even know how to approach it with the second half of it.

An object moves according to the equation x = vt + ke^(bt), where k, v, and b are constants, x represents distance in meters, t represents time in seconds, and e is the base of the natural logarithms.

a. Find the initial velocity (at time t=0)
b. Find the acceleration at t = 1/b seconds.

Homework Equations

x = vt + .5at²

The Attempt at a Solution

a: I assumed inital velocity would just be v... but the answer key says it's supposed to be v + kb.

b: Since the equation given was similar to the format of x = vt + .5at², I made the (questionable, I think) assumption that ke^bt = .5at².

Plugging in t = 1/b,

.5a(1/b²) = ke^b/b

a = 2keb²

This one was closer to the actual answer of keb², but still wrong and I'm very lost now.

I feel like I went in the completely wrong direction with this... any pointers?

 

[lewando]

The relevant equation you identified is for a system with constant acceleration. Not applicable here. If you are given an expression for x(t), what to you normally do to get the expression for v(t)? For a(t)?

 

[inaiki]

I think I'd take the derivative of x(t) for v(t), and the second derivative for a(t). But since v is already in the equation, does that still apply?

 

[PeroK]

I think I'd take the derivative of x(t) for v(t), and the second derivative for a(t). But since v is already in the equation, does that still apply?

I'd assume $v$ is supposed to be a constant here. As stated in the question, in fact. This means you are not going to be able to use the symbol $v$ for $dx/dt$.

 

[inaiki]

I'd assume $v$ is supposed to be a constant here. As stated in the question, in fact. This means you are not going to be able to use the symbol $v$ for $dx/dt$.

...Oh. I feel stupid now.

a. velocity = dx/dt = v + kbe^bt
when t=0,
dx/dt = v + kb

b. acceleration = dv/dt = kb²e^bt
when t = 1/b,
dv/dt = kb²e

Thank you!