A jet moves with a velocity (in $\text{m/s}$) of $v\left(t\right)=-e^{0.5t}+1.5t^2$. What is the jet's acceleration at $t=2.4s$?

• A. $3.88\text{ m/}{s}^2$
• B. $8.86\text{ m/}{s}^2$
• C. $7.20\text{ m/}{s}^2$
• D. $5.54\text{ m/}{s}^2$

Answer: (D)

Verified step by step guidance

1. Identify that acceleration is the time derivative of velocity, so we need to find the derivative of the given velocity function \(v(t) = -e^{0.5t} + 1.5t^2\) with respect to time \(t\).
2. Apply the differentiation rules: For the term \(-e^{0.5t}\), use the chain rule. The derivative of \(e^{0.5t}\) with respect to \(t\) is \$0.5 e^{0.5t}\(, so the derivative of \)-e^{0.5t}\( is \)-0.5 e^{0.5t}\(. For the term \)1.5t^2\(, use the power rule, which gives \)3.0 t$.
3. Write the expression for acceleration \(a(t)\) as the derivative of velocity: \(a(t) = \frac{d}{dt} v(t) = -0.5 e^{0.5t} + 3.0 t\).
4. Substitute the given time \(t = 2.4\) seconds into the acceleration expression to find the acceleration at that instant: \(a(2.4) = -0.5 e^{0.5 \times 2.4} + 3.0 \times 2.4\).
5. Evaluate the numerical values of the exponential and multiplication terms to get the acceleration value at \(t=2.4\) seconds.