Question

Two particles A and B are positioned as shown in
figure. If A and B start simultaneously towards origin
with respective velocitles 4 m/s and 3 m/s, then
closest distance between them is
-50 m
A 4 m/s
13 m/s
B-30 m​

Answer 1

closest distance between A and B should be 7.2 m

let after time t, particles A and B will be closest.

see figure,

distance covered by particle A , OA' = 50 - 4t

distance covered by particle B, OB' = 30 - 3t

now, ∆A'OB' is an right angled triangle.

so, distance between A and B , A'B' = √(OA'² + OB'²)

let A'B' = s

then, s = √{(50 - 4t)² + (30 - 3t)²}

squaring both sides,

s² = (50 - 4t)² + (30 - 3t)²

differentiating both sides,

2sds/dt = 2(50 - 4t)(-4) + 2(30 - 3t)(-3)

find value of t at ds/dt = 0.

⇒0 = -200 + 16t - 90 + 9t

⇒0 = -290 + 25t

⇒t = 290/25 = 54/5 = 10.8 sec

if you differentiate once again you will get, d²s/dt² > 0 so, at t = 10.8 sec, s will be minimum.

then, s = √{(50 -4 × 10.8)² + (30 - 3 ×10.8)²}

= 7.2 m

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