Question

A particle has a position Vector 3î-j^+2k^ at t=0.it moves with constant velocity (-î+j^-3k^)m/s.the position vector (in m) of the particle after 3 second is​

Answer 1

Answer:

• Position vector of particle after 3 sec = 2j -7k metres

Explanation:

Given:-

• A particle has position vector 3i - j + 2k at t= 0
• Particle moves with a constant velocity of (-i + j - 3k) m/s

To find:-

• Position vector (in m) of the particle after 3 seconds

Basic Knowledge required:-

• When the velocity of a body is constant its acceleration is zero.

• Dot product of any vector with a constant multiplies the dot product by the constant.

• Second equation of kinematics motion

        s = s₀ + v₀t + 1/2 at²

[ Where, s is position of body after a time t, s₀ is initial position of body, v₀ is initial velocity of body and a is acceleration of body ]

Solution:-

In this problem, since velocity of particle is constant therefore,

• acceleration of particle, a = 0
• time, t = 3 sec
• Initial position of particle, s₀ = 3i - j + 2k  
• Initial velocity of particle, v₀ = i + j - 3k  m/s

We need to find,

• the position of body after 3 sec, s =?

Using second equation of kinematics

→ s = s₀ + v₀t + 1/2 at²

→ s = (3i - j + 2k) + (-i + j -3k) (3) + 1/2 (0) (3)²

→ s = 3i - j + 2k - 3i + 3j - 9k + 0

→ s = 2j - 7k  m

Therefore,

• Position vector of particle after 3 seconds will be 2j - 7k metres.