Particle motion described by a time dependent function x(t)

[elizabethR]

how would i do this...
A particle moving along the x-axis has its position described by the function x=( 2.00 t^{3}- 5.00 t+ 1.00 )\; {\rm m}, where t is in s. At t= 4.00 , what are the particle's position- i calculated that to be 109m , velocity, and acceleration?

i have tried for over an hour on this one problem. i am clueless. my physics book sucks!

 

[Pyrrhus]

Show some work, so we can point out what's wrong.

Btw, for the position at t=4, you already have the x(t) (position function), for velocity and acceleration, look up the definitions.

$$v = \dot{x}$$

$$a = \dot{v} = \ddot{x}$$

 

[kyle8921]

I would like to first commend you for your persistence and effort in trying to solve this problem. Particle motion described by a time-dependent function is a common concept in physics and can be quite challenging to understand at first. However, with practice and understanding of the underlying principles, it becomes easier to solve such problems.

Now, let's break down the given function x(t) = 2.00t^3 - 5.00t + 1.00. This function represents the position of the particle along the x-axis at any given time t. The units for position are in meters (m). So, at t=4.00s, we can substitute this value into the function to find the particle's position: x(4.00) = 2.00(4.00)^3 - 5.00(4.00) + 1.00 = 109m.

This means that at t=4.00s, the particle's position is 109m along the x-axis. Now, to find the velocity and acceleration at this time, we need to take the first and second derivatives of the position function with respect to time, respectively.

Velocity (v) is the rate of change of position with respect to time, and it is given by the first derivative of the position function: v(t) = dx(t)/dt = 6.00t^2 - 5.00. Substituting t=4.00s, we get v(4.00) = 6.00(4.00)^2 - 5.00 = 91 m/s. Therefore, at t=4.00s, the particle's velocity is 91 m/s along the x-axis.

Acceleration (a) is the rate of change of velocity with respect to time, and it is given by the second derivative of the position function: a(t) = d^2x(t)/dt^2 = 12.00t. Substituting t=4.00s, we get a(4.00) = 12.00(4.00) = 48 m/s^2. Therefore, at t=4.00s, the particle's acceleration is 48 m/s^2 along the x-axis.

I hope this explanation helps you understand how to solve this type of problem. It's important to remember that practice makes perfect