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.