First slide

Projection Under uniform Acceleration

Question

A ship A is moving westwards with a speed of 10 km h^-1 and a ship B 100 km south of A, is moving northwards with a speed of 10kmh^-1 . The time after which the distance between them becomes shortest, is :

Moderate

A

5h
B

$5 \sqrt{2}$
C

$10 \sqrt{2}$
D

10

Solution

$|{\vec{v}}_{BA}| = \sqrt{10^{2} + 10^{2}}$
$= 10 \sqrt{2} kmph$
distance OB = 100 cos45°
$= 5 \sqrt{2} km$
Time taken to reach the shortest distance between A and B
$= \frac{50 \sqrt{2}}{|{\vec{v}}_{BA}|} = \frac{50 \sqrt{2}}{10 \sqrt{2}}$
t_sn = 5 hrs.

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