A dog running around a park has a velocity described by $v=(0.03t^2-2.5)\hat{ı}+(-0.1t+2)\hat{ȷ}$. What is the dog's displacement vector in unit-vector notation from $t=15s$ to $t=30s$?

• A.
• B.
• C.
• D.

Answer: (C)

Verified step by step guidance

1. Identify that the displacement vector \( \vec{r} \) from time \( t_1 \) to \( t_2 \) is the integral of the velocity vector \( \vec{v}(t) \) over that time interval: \[ \vec{r} = \int_{t_1}^{t_2} \vec{v}(t) \, dt \].
2. Write the velocity vector components explicitly: \[ \vec{v}(t) = \left(0.03t^2 - 2.5\right) \hat{\imath} + \left(-0.1t + 2\right) \hat{\jmath} \].
3. Set up the integrals for each component of displacement separately: \[ r_x = \int_{15}^{30} (0.03t^2 - 2.5) \, dt, \quad r_y = \int_{15}^{30} (-0.1t + 2) \, dt \].
4. Perform the integration for each component using the power rule for integrals: \[ \int t^n dt = \frac{t^{n+1}}{n+1} + C \], and evaluate the definite integrals by substituting the limits \( t=15 \) and \( t=30 \).
5. Combine the results of the two integrals to write the displacement vector in unit-vector notation: \[ \vec{r} = r_x \hat{\imath} + r_y \hat{\jmath} \].