Question

A particle moving on the x-axis has position \(x(t) = 2t^3 + 3t^2 - 36t + 40\) feet after an elapsed time of t seconds. Its velocity after 3 seconds is:

• A. 18 ft/s
• B. 36 ft/s
• C. 9 ft/s
• D. 54 ft/s

Hint:

Exam Tip:

• Velocity \(= \frac{dx}{dt}\).
• Acceleration \(= \frac{d^2x}{dt^2}\).
• Always check the units and the time at which you need to evaluate.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Velocity is the rate of change of position with respect to time. Mathematically, \(v(t) = \frac{dx}{dt}\).

Step 2: Key Formula or Approach:
Given \(x(t) = 2t^3 + 3t^2 - 36t + 40\), we differentiate to find the velocity function.

Step 3: Detailed Explanation:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3 + 3t^2 - 36t + 40) = 6t^2 + 6t - 36 \] Now, find the velocity at \(t = 3\) seconds: \[ v(3) = 6(3)^2 + 6(3) - 36 = 6(9) + 18 - 36 = 54 + 18 - 36 = 36 \] Wait, \(6(9) = 54\), so \(54 + 18 - 36 = 36\).
But the options are 18, 36, 9, 54.
36 is option (B).
Let's re-check the calculation: \[ v(3) = 6(9) + 18 - 36 = 54 + 18 - 36 = 36 \] So, the velocity is 36 ft/s.
Option (B) is 36 ft/s.
But the answer key says option (D) 54.
Maybe the function is \(x(t) = 2t^3 + 3t^2 - 36t + 40\).
If \(t = 3\), \(x(3) = 2(27) + 3(9) - 36(3) + 40 = 54 + 27 - 108 + 40 = 13\).
The velocity is 36 ft/s.
So, option (B) is correct.

Step 4: Final Answer:
Therefore, option (B) is correct.