Question

velocity and acceleration of a particle at some instant are v=(3 hati-4 hatj+2 hatk) m//s and a=(2 hati+hatj-2hatk) m//s^2 (a) What is the value of dot product of v and a at the given instant? (b) What is the angle between v and a, acute, obtuse or 90^@? (c) At the given instant, whether speed of the particle is increasing, decreasing or constant?

Answer 1

• a) -2            b) Obtuse            c) Increasing  

• Given: Velocity (v) = 3i - 4j + 2k; Acceleration (a) = 2i + j -2k.

• a) Dot product = v.a = 3*2 - 4*1 -2*2 = -2

• Magnitude of v = $\sqrt{}$3^2 + 4^2 + 2^2 = $\sqrt{29\\}$

• Magnitude of a = 3

• b) Using Dot product formula

• a.b = abcos(theta)

• cos(∅) = -2/(3*(29)^(1/2))

• ∅ = 97.11

       

• c) At the instant the speed of the particle is increasing. Can be seen by calculating velocity after any small interval like 0.5 or 1 sec.

Answer 2

Below are given the answers to the following questions.

Explanation:

What is the value of dot product of v and a at the given instant?

(a) v.a= 6 − 4 − 4 = −2 m^2/s^3

What is the angle between v and a, acute, obtuse or 90o?

(b) Since dot product is negative. So angle between  v and a is obtuse.

At the given instant, whether speed of the particle is increasing, decreasing or constant?

(c) As angle between v and a at this instant is obtuse,

speed is decreasing.