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The position of the particle is described by the function \( x(t) = 4.00 t^3 - 1.00 t + 3.00 \) (in meters), where \( t \) is in seconds. We need to calculate the particle's:

(a) **Position** at \( t = 3.00 \, \text{s} \)  
(b) **Velocity** at \( t = 3.00 \, \text{s} \)  
(c) **Acceleration** at \( t = 3.00 \, \text{s} \)

### Step 1: Position at \( t = 3.00 \, \text{s} \)
The position function is:
\[
x(t) = 4.00 t^3 - 1.00 t + 3.00
\]
Substitute \( t = 3.00 \) into the equation.

### Step 2: Velocity at \( t = 3.00 \, \text{s} \)
The velocity is the derivative of the position function with respect to time:
\[
v(t) = \frac{dx(t)}{dt} = \frac{d}{dt} \left( 4.00 t^3 - 1.00 t + 3.00 \right)
\]
We'll compute this derivative and evaluate it at \( t = 3.00 \, \text{s} \).

### Step 3: Acceleration at \( t = 3.00 \, \text{s} \)
The acceleration is the derivative of the velocity, or the second derivative of the position function:
\[
a(t) = \frac{dv(t)}{dt} = \frac{d^2 x(t)}{dt^2}
\]
We'll compute this second

A particle moving along the x-axis has its position described by the function x = (4.00 t^3 - 1.00 t + 3.00) m, where t is in seconds. At t = 3.00, what are the particle's (a) position, (b) velocity, and (c) acceleration?

Solve for the position, velocity, and acceleration of a particle moving along the x-axis with the function x(t) = 4.00 t^3 - 1.00 t + 3.00 at t = 3.00 seconds. This problem covers calculus concepts of differentiation and kinematics for high school students in grades 11-12.

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