this is the rest of the homework .thank you
Hello, carlasader!
Here's #7 . . .
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 m/sec,
where and are unit vectors directed due east and due north, respectively.
Initially A is at the origin O, and the position vector of B is .
At time seconds, the position vectors of A and B are and , resp.
a) Find expressions for and in terms of .
We have: .b) Hence write an expression for in terms is .
Then: .
Since B is always on a bearing of 45°, this occurs when:c) Find the time when the bearing of B from A is .
. . Then: . seconds.
The distance is: .d) Find the time when the cars are again 300 m apart.
If
Square both sides: .
Factor: .
Therefore, they are 300 m apart (again) when 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