First slide

Projection Under uniform Acceleration

Question

Two ships A and B are 10km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. Their distance of closest approach and how long do they take to reach it ?

Moderate

A

52 km, 15 min
B

5 2 km, 1.5 min
C

5 √2 km, 0.5 min
D

5√2 km, 15 sec

Solution

When a ship A is observing relative to ship B , Velocity of ship b relative to Ship A = $\large \vec V_{BA}$
we can write $\large \vec V_B = \vec V_A+\vec V_{BA}$
From the velocity triangle ,
$\large tan\theta=\frac{V_A}{V_B}=\frac{20}{20}=1 \Rightarrow \theta=45^0$
In triangle
x= shortest distance = Am=AB sin θ
.: x = 10x sin 45⁰ m = 5 √2 km
Now length BM = AB cosθ = 10 x cos 45⁰ m = 5√2 km
$\large \therefore t=\frac{BM}{V_{BA}}=\frac{BM}{\sqrt{V_A^2+V_B^2}}$
$\large =\frac{5\sqrt{2}}{\sqrt{20^2+20^2}}hour=15 min$

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