this is the rest of the homework .thank you
Hello, carlasader!
Here's #7 . . .
$\displaystyle \vec{r}\:=\:(20t)i$7. Two cars A and B are moving on straight horizontal roads with constant velocities.
The velocity of A is 20 m/sec due east, the velocity of B is $\displaystyle (10i + 10j)$ m/sec,
where $\displaystyle i$ and $\displaystyle j$ are unit vectors directed due east and due north, respectively.
Initially A is at the origin O, and the position vector of B is $\displaystyle 300i$.
At time $\displaystyle t$ seconds, the position vectors of A and B are $\displaystyle \vec{r}$ and $\displaystyle \vec{s}$, resp.
a) Find expressions for $\displaystyle \vec{r}$ and $\displaystyle \vec{s}$ in terms of $\displaystyle t$.
$\displaystyle \vec{s}\:=\:(300 + 10t)i + (10t)j$
We have: .$\displaystyle \vec{A} \:=\:\langle 20t,\,0\rangle,\;\vec{B}\:=\:\langle300 + 10t,\,10t\rangle$b) Hence write an expression for $\displaystyle \overrightarrow{AB}$ in terms is $\displaystyle t$.
Then: .$\displaystyle \overrightarrow{AB} \;=\;\langle 300-10t,\:10t\rangle$
Since B is always on a bearing of 45°, this occurs when: $\displaystyle A \,= \,\langle 300t,\:0\rangle$c) Find the time when the bearing of B from A is $\displaystyle 045^o$.
. . Then: .$\displaystyle \langle20t,\:0\rangle \:=\:\langle 300,\:0\rangle\quad\Rightarrow\quad t = 15$ seconds.
The distance is: .$\displaystyle |AB| \;=\;\sqrt{(300 - 10t)^2 + (10t-0)^2} \;=\;\sqrt{90,000 - 6000t + 200t^2}$d) Find the time when the cars are again 300 m apart.
If $\displaystyle |AB| = 300:\;\;\sqrt{90,000 - 6000t + 200t^2} \:=\:300$
Square both sides: .$\displaystyle 90,000 - 6000t + 200t^2\:=\:90,000\quad\Rightarrow\quad 200t^2 - 6000t \:=\:0$
Factor: .$\displaystyle 200t(t - 30) \:=\:0\quad\Rightarrow\quad t \,=\,0,\:30$
Therefore, they are 300 m apart (again) when $\displaystyle t = 30$ seconds.
a) As the forces acting on B are -3mg and T, and the accelleration of the
particle is -(1/2)g, the tension must be (1/2)3mg. That is we have:
-3mg + T = 3ma,
so:
T = 3ma + 3mg
but a= -(1/2)g.
b) Resolve the gravitational force on A into components up the plene and
normal to the plane.
The component normal to the plane is -mg sin(theta), and up the plane is
-mg cos(theta), also sin(theat)=3/5, and cos(theta)=4/5.
Now the forces on the particle up the plane are T = (3/2)mg, the component
of the weight acting up the plane -(4/5)mg, and the friction -mu (3/5)mg
So the total force up the plane is:
T - (4/5)mg - mu (3/5)mg
and the acceleration up the plane is (1/2)g, so:
(3/2)mg - (4/5)mg - mu (3/5)mg = (1/2) mg
or cancelling mg:
(3/2) - (4/5) - mu (3/5) = (1/2)
Which I will leave to the reader to finish.
RonL