Question

two particles have position vector (4i^-4j^+7k^)and(2i^+2j^+5k^)m representatively.if the velocity of first particle is 0.4ms -1then find the velocity of second particle when they collide after 10 seconds​

Answer 1

Answer:

6.39 m/s

Step-by-step explanation:

Let a = 4i + 4j + 7k and,

b = 2i + 2j + 5k

Hence,

$|a| = \sqrt{16 + 16 + 49} = \sqrt{81} \\ |b| = \sqrt{4 + 4 + 25} = \sqrt{33}$

The time taken is same for both the velocities.

Therefore,

$\frac{ \sqrt{81} }{10} = \frac{ \sqrt{33} }{v}$

We get,

$v = 6.39$

Hence, the velocity of the 2nd particle is approximately 6.39 m/s.