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Object A has a position as a function of time given by A(t) = (3.00 m/s)t i^ + (1.00 m/s2)t2 j^. Object B has a position as a function of time given by B(t) = (4.00 m/s)t i^ + (−1.00 m/s2)t2 j^. All quantities are SI units. What is the distance between object A and object B at time t = 5.00 s?

Respuesta :

Answer:50.25 m

Explanation:

Given

Object A has a position given by

[tex]A(t)=3t\hat{i}1\cdot t^2\hat{j}[/tex]

Object B position is given by

[tex]B(t)=4t\hat{i}+(-1)t^2\hat{j}[/tex]

at [tex]t=5\ s[/tex]

[tex]A(5)=15\hat{i}+25\hat{j}[/tex]

[tex]B(5)=20\hat{i}-25\hat{j}[/tex]

Distance between A and B is given by the magnitude of vector r

[tex]\vec{r}=5\hat{i}-50\hat{j}[/tex]

[tex]|\vec{r}|=\sqrt{5^2+(-50)^2}[/tex]

[tex]|\vec{r}|=50.25\ m[/tex]  

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