# A level Maths 05- cont.

• May 23rd 2007, 06:53 AM
A level Maths 05- cont.
this is the rest of the homework .thank you:p
• May 25th 2007, 06:05 PM
Soroban

Here's #7 . . .

Quote:

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 $(10i + 10j)$ m/sec,
where $i$ and $j$ are unit vectors directed due east and due north, respectively.
Initially A is at the origin O, and the position vector of B is $300i$.
At time $t$ seconds, the position vectors of A and B are $\vec{r}$ and $\vec{s}$, resp.

a) Find expressions for $\vec{r}$ and $\vec{s}$ in terms of $t$.

$\vec{r}\:=\:(20t)i$
$\vec{s}\:=\:(300 + 10t)i + (10t)j$

Quote:

b) Hence write an expression for $\overrightarrow{AB}$ in terms is $t$.
We have: . $\vec{A} \:=\:\langle 20t,\,0\rangle,\;\vec{B}\:=\:\langle300 + 10t,\,10t\rangle$

Then: . $\overrightarrow{AB} \;=\;\langle 300-10t,\:10t\rangle$

Quote:

c) Find the time when the bearing of B from A is $045^o$.
Since B is always on a bearing of 45°, this occurs when: $A \,= \,\langle 300t,\:0\rangle$
. . Then: . $\langle20t,\:0\rangle \:=\:\langle 300,\:0\rangle\quad\Rightarrow\quad t = 15$ seconds.

Quote:

d) Find the time when the cars are again 300 m apart.
The distance is: . $|AB| \;=\;\sqrt{(300 - 10t)^2 + (10t-0)^2} \;=\;\sqrt{90,000 - 6000t + 200t^2}$

If $|AB| = 300:\;\;\sqrt{90,000 - 6000t + 200t^2} \:=\:300$

Square both sides: . $90,000 - 6000t + 200t^2\:=\:90,000\quad\Rightarrow\quad 200t^2 - 6000t \:=\:0$

Factor: . $200t(t - 30) \:=\:0\quad\Rightarrow\quad t \,=\,0,\:30$

Therefore, they are 300 m apart (again) when $t = 30$ seconds.

• May 25th 2007, 09:54 PM
CaptainBlack
http://www.mathhelpforum.com/math-he...untitled-5.jpg

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