The position of a small object is given by $x=0.6e^{2t}-1$, where $t$ is in seconds and $x$ in meters. At what time is the small object travelling with a velocity $v=65.5\text{ m/s}$?

• A. $2.35\text{ s}$
• B. $2.0\text{ s}$
• C. $2.69\text{ s}$
• D. Never

Answer: (B)

Verified step by step guidance

1. Identify the given position function: \(x = 0.6 e^{2t} - 1\), where \(t\) is time in seconds and \(x\) is position in meters.
2. Recall that velocity \(v\) is the time derivative of position \(x\), so find \(v(t)\) by differentiating \(x(t)\) with respect to \(t\): \(v = \frac{dx}{dt}\).
3. Differentiate \(x = 0.6 e^{2t} - 1\) using the chain rule: the derivative of \(e^{2t}\) is \$2 e^{2t}$, so \(v = 0.6 \times 2 e^{2t}\).
4. Set the velocity equal to the given value \(v = 65.5\) m/s and solve the equation \$1.2 e^{2t} = 65.5\( for \)t$.
5. Take the natural logarithm of both sides to isolate \(t\): \(2t = \ln\left(\frac{65.5}{1.2}\right)\), then solve for \(t\) by dividing both sides by 2.