Question

The displacement of a particle at time \( t \) is given by \( s = 2t^3 - 5t^2 + 4t - 3 \). Find the velocity and displacement at the time when the acceleration is \( 14 \, \text{ft/sec}^2 \).

Hint:

To find velocity and displacement, first find the first and second derivatives of the displacement function, then substitute the given time into these equations.

Answer 1

Step 1: Find the velocity and acceleration.
The velocity is the first derivative of the displacement: \[ v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 5t^2 + 4t - 3) = 6t^2 - 10t + 4 \] The acceleration is the derivative of the velocity: \[ a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 10t + 4) = 12t - 10 \]

Step 2: Find the time when acceleration is 14 \( \text{ft/sec}^2 \).
Set the acceleration equal to 14 and solve for \( t \): \[ 12t - 10 = 14 \] \[ 12t = 24 $\Rightarrow$ t = 2 \]

Step 3: Find the velocity and displacement at \( t = 2 \).
Substitute \( t = 2 \) into the velocity equation: \[ v = 6(2)^2 - 10(2) + 4 = 24 - 20 + 4 = 8 \, \text{ft/sec} \] Substitute \( t = 2 \) into the displacement equation: \[ s = 2(2)^3 - 5(2)^2 + 4(2) - 3 = 16 - 20 + 8 - 3 = 1 \, \text{ft} \]

Final Answer: At \( t = 2 \), the velocity is \( \boxed{8 \, \text{ft/sec}} \) and the displacement is \( \boxed{1 \, \text{ft}} \).