A toy airplane is flying in an $\text{x/y}$ plane parallel to the ground with a velocity $v\left(t\right)=(4t-0.015t^3)\hat{ı}+(1.8+0.5t)\hat{ȷ}$. What is the magnitude of the airplane's acceleration at $t=30s$?

• A. $36.5\text{ m/}{s}^2$
• B. $285\text{ m/}{s}^2$
• C. $-36.0\text{ m/}{s}^2$
• D. $-268.2\text{ m/}{s}^2$

Answer: (A)

Verified step by step guidance

1. Identify the velocity vector function given as \(\mathbf{v}(t) = \left(4t - 0.015t^3\right)\hat{i} + \left(1.8 + 0.5t\right)\hat{j}\). This represents the velocity components in the x and y directions as functions of time.
2. Recall that acceleration is the time derivative of velocity. Therefore, find the acceleration vector \(\mathbf{a}(t)\) by differentiating each component of \(\mathbf{v}(t)\) with respect to time \(t\): \(\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t) = \frac{d}{dt} \left[ \left(4t - 0.015t^3\right)\hat{i} + \left(1.8 + 0.5t\right)\hat{j} \right]\).
3. Calculate the derivatives of each component separately: For the \(\hat{i}\) component: \(a_x(t) = \frac{d}{dt} \left(4t - 0.015t^3\right)\), For the \(\hat{j}\) component: \(a_y(t) = \frac{d}{dt} \left(1.8 + 0.5t\right)\).
4. Evaluate the acceleration components at the given time \(t = 30\) seconds by substituting \(t=30\) into the expressions for \(a_x(t)\) and \(a_y(t)\).
5. Find the magnitude of the acceleration vector at \(t=30\) seconds using the Pythagorean theorem: \(|\mathbf{a}(30)| = \sqrt{a_x(30)^2 + a_y(30)^2}\). This gives the total acceleration magnitude of the airplane at that instant.