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Magnitude of Instantaneous Velocity

Point A moves...

Question

Point A moves uniformly with velocity v so that the vector v is continually "aimed" at point B which in its turn moves rectilinearly and uniformly with velocity u < v. At the initial moment of time $v ⊥ u$ and the points are separated by a distance I.How soon will the points converge ?

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Solution

As two velocities are perpendicular, relative velocity of one with respect to other is, $\sqrt{u^{2} + v^{2}}$

So time taken to converge is, $t = \frac{l}{\sqrt{v^{2} + u^{2}}}$

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Point $A$ moves uniformly with velocity $v = 13 m/s$ so that the vector $u = 5 m/s$ is continually aimed at point $B$ which in turn moves rectilinearly and uniformly with velocity $u < v$ . At the initial moment of time, $v ⊥ u$ and the points are separated by a distance $l = 12 m$ . How soon will the points converge?

A

moves uniformly with velocity

v

so that the vector

v

is continually "aimed" at point

B

which in its turn moves rectilinearly and uniformly with velocity

u < v

. At the initial moment of time

v ⊥ u

and the points are separated by a distance

l

. The time at which the points converge is given as

T = \frac{x v l}{v^{2} - u^{2}}

x

Q. Particle A moves uniformly with velocity v, so that vector v is continually aimed at point B which moves rectilinearly with a velocity u < v. At t = 0, v and u are perpendicular. Find the time when they converge. Assume A and B are separated by

l

A

moves uniformly with velocity

v

v

is continually aimed at point

B

which moves rectilinearly with a velocity

u < v

t = 0

v

u

are perpendicular. The time when they converge is given as

t = \frac{x l v}{v^{2} - u^{2}}

x

A

B

are separated by

l

t = 0

^{'} A^{'}

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{\to v}_{1}

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v_{2} (= 10 m / s e c)

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as shown in the figure. If the time (in sec)when the points A and B converge is

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