time and distance ..please help with steps.

If A and B are running in a circular track in a direction opposite to that in which C is running, who is running at twice and thrice the speeds of A and B respectively and on the same track. They start running from a same point. It is known that A's average speed is 3 m/s and the track is 120 m.

When after the start, will B find himself equdistant and between A and C for the first time?

Re: time and distance ..please help with steps.

Given the speed of A, find the speed of B and C.

The circular track has 2*pi radians and 120 meters long so people running round it change angle at a rate of 2*pi/120= pi/60 radians per meter.

If B is running with speed v then he is changing direction at a rate of v*(pi/60) radians per second.

After a time t B will have gone v*t*(pi/60) radians

Putting the circle in the x-y plane B's x co-ordinate is r*cos(v*t*(pi/60)) where r is the radius of the circle

B's y co-ordinate will be r*sin(v*t*(pi/60))

Likewise if A is running at speed u then his x and y co-ordinates will be r*cos(u*t*(pi/60)) and r*sin(u*t*(pi/60))

C is running in the opposite direction at a speed w. Because its going the opposite direction you have to put a minus sign in front, so C's x co-ordinate is -r*cos(w*t*(pi/60)) and y co-ordinate is -r*sin(w*t*(pi/60))

Now you have their co-ordinates in terms of time get an expression for the distance between their points.

Distance |BC|= |AB|

That gives you one equation in terms of time, solve the equation for t.

Remember all angles are in radians.

Re: time and distance ..please help with steps.

Is there a way to solve by just using algebra?

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Re: time and distance ..please help with steps.

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I am sure you would understand the algebraic method and also notice how making a drawing makes things easy to understand.

Re: time and distance ..please help with steps.

Hello, hisajesh!

Quote:

$\displaystyle A\text{ and }B\text{ are running in a circular track in a direction opposite to that in which }C\text{ is running,}$

$\displaystyle \text{who is running at twice and thrice the speeds of }A\text{ and }B\text{, respectively, and on the same track.}$

$\displaystyle \text{They start running from the same point at the same time.}$

$\displaystyle \text{It is known that }A[\text{'s average speed is 3 m/s and the track is 120 m long.}$

$\displaystyle \text{When after the start, will }B\text{ find himself equdistant and between }A\text{ and }C\text{ for the first time?}$

We find that: .$\displaystyle \begin{Bmatrix}\text{A's speed}&=& \text{3 m/s} \\ \text{B's speed} &=& \text{2 m/s} \\ \text{C's speed}&=& \text{6 m/s} \end{Bmatrix}$

Picture the circular track; circumference 120 meters.

The one-meter marks are labeled clockwise: 1, 2, 3 . . . ending with 120 at the top.

A and B are running clockwise; C is running counter-clockwise, starting at the top.

In $\displaystyle t$ seconds, $\displaystyle A$ has run to point $\displaystyle 3t.$

In $\displaystyle t$ seconds, $\displaystyle B$ has run to point $\displaystyle 2t.$

In $\displaystyle t$ seconds, $\displaystyle C$ has run to point $\displaystyle 120-6t.$

When is $\displaystyle B$ exactly halfway between $\displaystyle A$ and $\displaystyle C$?

When is $\displaystyle B$'s coordinate the *average* of $\displaystyle A$'s coordinate and $\displaystyle C$'s coordinate?

We have: .$\displaystyle 2t \:=\:\frac{3t + (120-6t)}{2} \quad\Rightarrow\quad 4t \:=\:120-3t \quad\Rightarrow\quad 7t \,=\,120$

Therefore: .$\displaystyle t \:=\:\frac{120}{7} \:=\:17\tfrac{1}{7}\text{ seconds.}$