Question

A particle has, a velocity U= 3i+ 4j+12k m/s in a co-ordinate system moving with a velocity 0.8c relative to laboratory along positive direction of X- axis .Find u in a laboratory frame ​

Answer 1

Answer:

If r is the position vector of the particle at t=0 then

dr/dt)°=3i+4j

and acceleration

d/dt(dr/dt)=0.4i+ 0.3j

Integrating wrt t we get

dr/dt=(0.4i+ 0.3j)t + c

Here c is the integration constant.

Putting at t=0

dr/dt)°=3i+4j

we get

C=3i+4j

So at time t the velocity of the particle is

dr/dt=(0.4i +0.3) t+ 3i+4j

at t=10s , velocity is

V=dr/dt=7i+ 7j

And

|V|=7√2 m/s

Answer 2

Given:

U = 3i+4j+12k m/s

V = 0.8c

To Find:

u

Solution:

Let r be the positive vector of the particle at t = 0.

Then,

dr/dt = 3i + 4

acceleration is d/dt(dr/dt) = 0.4i + 0.3,

dr/dt = (0.4i + 0.3j)t + c

(Here, c is the integration constant)

Putting t = 0

dr/dt = 3i + 4j

C = 3i + 4j

The velocity of particle is

dr/dt = (0.4i + 0.3)t + 3i + 4j

at time t = 10s

V = dr/dt

   = 7i+7j

|V| = 7$\sqrt{2}$m/s.

Therefore, u in a laboratory frame = 7$\sqrt{2}$m/s.