2019-06-03T04:41:14-04:00
A rabbit run across a parking lot. The coordinate of the rabbits as a function of time t are x=0.3t square +7.2t square+28 Y=0.22t square-9.1t+80 with the in seconds and x and y in meters. At the =15s, what is the rabbits position vector r in unit vector notation and as magnitude and an angle
1
2019-06-03T10:41:16-0400
x = 0.3 t 2 + 7.2 t 2 + 28 = 7.5 t 2 + 28 y = 0.22 t 2 − 9.1 t + 80 x= 0.3t^2+7.2t^2+28 = 7.5t^2+28\\
y=0.22t^2-9.1t+80 x = 0.3 t 2 + 7.2 t 2 + 28 = 7.5 t 2 + 28 y = 0.22 t 2 − 9.1 t + 80
r ⃗ ( t ) = x ( t ) ⋅ i ⃗ + y ( t ) ⋅ j ⃗ r ⃗ ( 15 ) = ( 7 , 5 ⋅ 1 5 2 + 28 ) ⋅ i ⃗ + ( 0 , 22 ⋅ 1 5 2 − 9.1 ⋅ 15 + 80 ) j ⃗ = = 1715.5 ⋅ i ⃗ − 7 ⋅ j ⃗ \vec{r}(t)=x(t) \cdot \vec{i} +y(t)\cdot \vec{j}\\
\vec{r}(15) = (7,5\cdot 15^2+28)\cdot\vec{i}+(0,22\cdot 15^2-9.1 \cdot 15 +80)\vec{j}=\\
= 1715.5 \cdot\vec{i}-7\cdot\vec{j} r ( t ) = x ( t ) ⋅ i + y ( t ) ⋅ j r ( 15 ) = ( 7 , 5 ⋅ 1 5 2 + 28 ) ⋅ i + ( 0 , 22 ⋅ 1 5 2 − 9.1 ⋅ 15 + 80 ) j = = 1715.5 ⋅ i − 7 ⋅ j
∣ r ⃗ ( 15 ) ∣ = 1715. 5 2 + 7 2 = 1715.51 m ϕ ( 15 ) = t a n − 1 ( y ( 15 ) x ( 15 ) ) = t a n − 1 ( − 7 1715.5 ) = − 0.23 d e g \lvert \vec{r}(15) \rvert = \sqrt{1715.5^2+7^2} = 1715.51 m \\
\phi(15) = tan^{-1}(\frac{y(15)}{x(15)}) = tan^{-1}(\frac{-7}{1715.5}) = -0.23 deg ∣ r ( 15 )∣ = 1715. 5 2 + 7 2 = 1715.51 m ϕ ( 15 ) = t a n − 1 ( x ( 15 ) y ( 15 ) ) = t a n − 1 ( 1715.5 − 7 ) = − 0.23 d e g
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS !
Comments