Understanding Mosquito Acceleration from Velocity

The problem asks for the magnitude of a mosquito's acceleration, given its velocity vector as a function of time. Acceleration is defined as the rate of change of velocity with respect to time. We can find the acceleration vector by differentiating the velocity vector with respect to time.

Calculating the Acceleration Vector

The given velocity vector is:

$$ \vec{v} = 5\hat{i} + 3t\hat{j} + 2t^2\hat{k} $$

To find the acceleration vector, $\vec{a}$, we differentiate $\vec{v}$ with respect to time ($t$):

$$ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(5\hat{i} + 3t\hat{j} + 2t^2\hat{k}) $$

We differentiate each component separately:

• The derivative of the $\hat{i}$ component is $\frac{d}{dt}(5) = 0$.
• The derivative of the $\hat{j}$ component is $\frac{d}{dt}(3t) = 3$.
• The derivative of the $\hat{k}$ component is $\frac{d}{dt}(2t^2) = 2 \times 2t = 4t$.

So, the acceleration vector is:

$$ \vec{a} = 0\hat{i} + 3\hat{j} + 4t\hat{k} $$

$$ \vec{a} = 3\hat{j} + 4t\hat{k} $$

Determining the Magnitude of Acceleration

The magnitude of a vector $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ is given by the formula $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$.

For our acceleration vector $\vec{a} = 0\hat{i} + 3\hat{j} + 4t\hat{k}$, the components are $a_x = 0$, $a_y = 3$, and $a_z = 4t$. Therefore, the magnitude of the acceleration is:

$$ |\vec{a}| = \sqrt{0^2 + 3^2 + (4t)^2} $$

$$ |\vec{a}| = \sqrt{0 + 9 + 16t^2} $$

$$ |\vec{a}| = \sqrt{9 + 16t^2} $$

Finding the Magnitude Corresponding to the Options

The calculated magnitude $|\vec{a}| = \sqrt{9 + 16t^2}$ depends on time ($t$). However, the options provided are constant values. This suggests that the question might be asking for the magnitude at a specific time, or there might be an intended time value that yields one of the options.

Let's test the options. If we want the magnitude to be $5 \text{ m/s}^2$, we can set our calculated magnitude equal to 5:

$$ \sqrt{9 + 16t^2} = 5 $$

Squaring both sides:

$$ 9 + 16t^2 = 5^2 $$

$$ 9 + 16t^2 = 25 $$

Subtracting 9 from both sides:

$$ 16t^2 = 25 - 9 $$

$$ 16t^2 = 16 $$

Dividing by 16:

$$ t^2 = 1 $$

This gives $t = 1$ second (assuming time $t \ge 0$).

Thus, at time $t=1$, the magnitude of the mosquito's acceleration is $5 \text{ m/s}^2$. Since $5 \text{ m/s}^2$ is one of the options, this is the intended answer.

Final Answer

The magnitude of the mosquito's acceleration is $5 \text{ m/s}^2$ (specifically at $t=1$ second).