The position function for a particle moving on the $x$-axis is $x\left(t\right)=-2t^2+18t+7$. What is the particle's initial velocity?

• A. $v_0=18\mathrm{m/s}$
• B. $v_0=7\mathrm{m/s}$
• C. $v_0=9\mathrm{m/s}$
• D. $v_0=4.5\mathrm{m/s}$

Answer: (A)

Verified step by step guidance

1. Identify the given position function of the particle: \(x(t) = -2t^2 + 18t + 7\).
2. Recall that velocity is the time derivative of position, so write the velocity function as \(v(t) = \frac{dx}{dt}\).
3. Differentiate the position function term-by-term: the derivative of \(-2t^2\) is \(-4t\), the derivative of \$18t$ is \(18\), and the derivative of the constant \(7\) is \(0\).
4. Combine these results to express the velocity function: \(v(t) = -4t + 18\).
5. To find the initial velocity, evaluate the velocity function at \(t=0\): \(v_0 = v(0) = -4(0) + 18\).