Question

Object A has a position as a function of time given by rA(t) = (3.00 m/s)t i ^ + (1.00 m/s2)t2j^. Object B has a position as a function of time given byrB(t) = (4.00 m/s)ti^ + (-1.00 m/s2)t2j^. All quantities are SI units. What is the distance between object A and object B at time t = 3.00 s?A) 3.46 m B) 15.0 m C) 18.3 m D) 34.6 m E) 29.8 m

Answer 1

Option C is the correct answer.

Explanation:

Given that

             $rA(t)=(3.00 m/s)t\hat{i}+ (1.00 m/s^2)t^2\hat{j}\texttt{ and }rB(t)=(4.00 m/s)t\hat{i}+ (-1.00 m/s^2)t^2\hat{j}$

We need to find distance when t = 3 s

Substituting t = 3 s

             $rA(t)=(3.00 m/s)\times 3\hat{i}+ (1.00 m/s^2)\times 3^2\hat{j}=9\hat{i}+9\hat{j}\\\\rB(t)=(4.00 m/s)\times 3\hat{i}+ (-1.00 m/s^2)\times 3^2\hat{j}=12\hat{i}-9\hat{j}$

$\texttt{Displacement = }12\hat{i}-9\hat{j}-(9\hat{i}+9\hat{j})=3\hat{i}-18\hat{j}$

$\texttt{Magnitude = }\sqrt{3^2+(-18)^2}=18.3m$

Option C is the correct answer.

Answer 2

The distance between object A and object B is approximately 18.248 meters. (Choice C)

How to calculate the distance between two objects

In this question we must apply the concepts of vector difference, dot product and norm to determine the distance between objects A and B, in meters:

$r_{B/A} = \sqrt{(\vec r_{B}-\vec r_{A})\,\bullet\,(\vec r_{B}-\vec r_{A})}$ (1)

Where:

• $\vec r_{A}$ - Vector distance of object A, in meters.
• $\vec r_{B}$ - Vector distance of object B, in meters.
• $r_{B/A}$ - Distance of B relative to A, in meters.

If we know that $\vec r_{A} = (3\cdot t, t^{2})\,\left[m\right]$, $\vec r_{B} = (4\cdot t,-t^{2})\,\left[m\right]$ and $t = 3\,s$, then the distance of B relative to A is:

$r_{B/A}=\sqrt{t^{2}+4\cdot t^{4}}$

$r_{B/A} = t\cdot \sqrt{1+4\cdot t^{2}}$

$r_{B/A} = 3\cdot \sqrt{1+4\cdot 3^{2}}$

$r_{B/A} \approx 18.248\,m$

The distance between object A and object B is approximately 18.248 meters. (Choice C) $\blacksquare$

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