Question

A particle is kept at rest at origin. Another particle starts from (5m, 0) with a velocity of - 4 icap + 3jcap
Find their closest distance of approach.
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Answer 1

Answer:

ans should be 3m......

After time t position of particle will be (5m−4t)iˆ+3tjˆ

So distance from the distance from the particle at origin is given by

 D = root under(5m−4t)^2+(3t)^2

Now for displacement to be minimum, 

dD/dt have to be zero, hence we have

d/dt root under(5m−4t)^2+(3t)^2 =0

⇒18t−8(5m−4t)/2×root under(5m−4t)^2+(3t)^2=0

⇒t=4/5

Now putting the value of t in the equation of distance we get, D=3. Hence the minimum of approach will be 3. 

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Answer 2

Answer:

The two particle's closest distance of approach is 3m.

Explanation:

Given, one particle is kept at rest at origin while the other particle is moving.

At time t, the position of moving particle is,

D = $(5-4t)\hat{i} + 3t\hat{j}$

Distance of this particle from the particle at origin is,

D = $\sqrt{(5-4t)^{2} + 3t^{2} }$..............................................................(1)

To find the closest distance of approach displacement should be minimum.

For displacement to be minimum $\frac{dD}{dt}$ have to be zero, it is given as,

$\frac{dD}{dt} = 0$

⇒ $\frac{50t - 40}{2\sqrt{(5-4t)^{2}+3t^{2} } } =0$

⇒ $t=\frac{4}{5} s$

Now, substitute the value of t in equation(1)

$D=\sqrt{(5-4\times\frac{4}{5} )^{2}+(3\times\frac{4}{5}) ^{2} }$

⇒ $D=3m$

Therefore, the closest distance of approach is 3m.

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