Question

I need help with this question 1,2 and 3 they relate

Answer 1

1. 15 meters

2. 11.629 m/s

3. 334.54°

Explanation:

1.

We have the following functions:

`\begin{gathered} x=-2+3t+1.5t^2 \\ y=4-t^2 \end{gathered}`

We have that the radius vector is defined as follows:

`\begin{gathered} r=xi+yj \\ \text{ replacing x and y:} \\ r=(-2+3t+1.5t^2)i+(4-t^2)j \\ \text{ now, replacing t = 2.5} \\ r=(-2+3\cdot2.5+1.5(2.5)^2)i+(4-(2.5^2))j \\ r=14.875i-2.25j \\ |r|=\sqrt[]{(1.875)^2+(-2.25)^2} \\ |r|=15.04\cong15\text{ m} \end{gathered}`

2.

The velocity is given as follows:

`\begin{gathered} v=(dr)/(\differentialDt t) \\ \text{replacing:} \\ v=(d)/(dt)((-2+3t+1.5t^2)i+(4-t^2)j) \\ v=(3+3t)i+(-2t)j \\ t=2.5,\text{ replacing:} \\ v=(3+3\cdot2.5)i+(-2\cdot2.5)j \\ v=10.5i-5j \\ |v|=\sqrt[]{(10.5)^2+(-5)^2} \\ |v|=11.629\text{ m/s} \end{gathered}`

3.

Now, to calculate the angle, we do it as follows:

`\begin{gathered} \tan \theta=(-5)/(10.5) \\ \theta=\arctan \mleft((-5)/(10.5)\mright) \\ \theta=-25.46\text{\degree} \\ \text{ therefore,} \\ 360\text{\degree}-25.46\text{\degree}=334.54\text{\degree} \end{gathered}`