Need help with a Very Hard Kinematics Qn

physicsnoob93 · Jan 13, 2009

1. The problem statement, all variables and given/known data

Imagine a tank traveling at constant speed v1 along direction AB and a missile is chasing the tank with constant speed v2. The direction of motion of the missile is always pointing toward the tank. At the t the tank is at position F and the missile is at position D, where FD is perpendicular to AB and FD = L. What is the acceleration of the missile at this instant?

A---------------F---------------B
************|
************|
************|
************|
************D

I have no clue, i have spent hours on this qn. I tried drawing the initial and final velocity vectors for the missile but i have no idea how to continue from there.

Thanks if anyone can help.

2. Relevant equations
?

3. The attempt at a solution

chrisk · Jan 13, 2009

Acceleration is a change in velocity with respect to time. The missle velocity relative to the tank is v2 - v1. There is no linear accleration in this problem. Are you trying to solve for the angular acceleration?

Thaakisfox · Jan 13, 2009

Its best to use relative coordinates.
Lets fix our system to the tank, and use polar coordinates, so r will be the instantaneous distance between the missile and the tank, and \phi will be the angle between this and the initial position.
Our equations are then:

[tex]\dot r=v_2-v_1\sin\varphi[/tex]

and

[tex]r\dot\varphi=v_1\cos\varphi[/tex]

Now we know that the acceleration in polar coordinates is:

[tex]\vec a = (\ddot r -r\dot\varphi^2)\vec e_{r} + (2\dot r\dot\varphi + r\ddot\varphi)\vec e_{\varphi}[/tex]

Using the equations, calculate the acceleration, and after you are done put [tex]\varphi=0[/tex] and r=L, and you will get it at the beginning.

It turns out to be:

[tex]a=\frac{v_1v_2}{L}[/tex]

(I be wrong, because I just scribbled it up quickly... :D)

physicsnoob93 · Jan 14, 2009

Thanks for the replies. I solved it using similar triangles and i got v1v2/L too.

My Solution is:

If theres a triangle drawn for the distance,

Theres the adjacent with l and the opposite to be V1*dt

If theres a triangle for velocity,
The adjacent is V2, the opposite is a*dt
Then we get

V1dt/l = adt/V2
Simplifying,
a = V1V2/l

Thanks though :D

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Need help with a Very Hard Kinematics Qn

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