Acceleration of the particle at t = 1s

[patelneel1994]

Homework Statement

The position of a particle as it moves along the x-axis is given by x = 15e-2t m, where t is in s. What is the acceleration of the particle at t = 1 s?

Homework Equations

d = vi.t + 1/2 a ts
since I don't know how distance is implemented.

The Attempt at a Solution

I couldn't do it
= 15e-2t = 0 + 1/2 a 1 s2

 

[hjelmgart]

Can't you find a value for x with the equation you have? That should be all you need!

 

[collinsmark]

Homework Statement

The position of a particle as it moves along the x-axis is given by x = 15e-2t m, where t is in s. What is the acceleration of the particle at t = 1 s?

Homework Equations

d = vi.t + 1/2 a ts
since I don't know how distance is implemented.

The Attempt at a Solution

I couldn't do it

Per the forum rules, you must at least try and show your effort before help is allowed.

= 15e-2t = 0 + 1/2 a 1 s2

And at the very least, please put some effort into making it easy to read what the problem actually is. Do you really mean that

x = 15e - 2t [m]?

Or do you mean,

x = 15e^-2t [m]?

Or is it,

x = 0.15t [m]?

One is much more likely to get help if one does the due diligence to make the original post easy to read.

 

[haruspex]

The position of a particle as it moves along the x-axis is given by x = 15e-2t m, where t is in s. What is the acceleration of the particle at t = 1 s?

If you are given the position as a function of time, x = f(t), what operation do you need to perform to find the velocity function? And what to get the acceleration function?

 

[patelneel1994]

To find the velocity = delta x/ delta t

 

[collinsmark]

What operation does $\frac{\Delta x}{\Delta t}$ become as $\Delta t$ and the corresponding $\Delta x$ become smaller and smaller to the point of being infinitesimally small?

In other words, if you graph x(t) vs. t (x on the vertical axis and t on the horizontal axis), what is the slope of the curve at any point in time, t?