Two particles are fired together along a smooth horizontal surface with velocities 4 m/s and 5 m/s. If they move at 60° to each other, find the distance between them in 2 seconds.
-
A.
(2sqrt{61}) -
B.
(sqrt{42}) -
C.
(2sqrt{21}) -
D.
(2sqrt{10})
Correct Answer: Option C
Explanation
Given lines (OA) and (OB) inclined at angle (theta), the line (AB) is gotten using cosine rule.
(|AB|^{2} = |OA|^{2} + |OB|^{2} – 2|OA||OB|cos theta)
(|AB|^{2} = 4^{2} + 5^{2} – 2(4)(5)cos 60)
= (16 + 25 – 20)
(|AB|^{2} = 21 implies |AB| = sqrt{21})
(implies text{The two particles are} sqrt{21} m text{apart in 1 sec})
In two seconds, the particles will be (2sqrt{21} m) apart.