Question

The velocity v of a moving particle changes with
displacement x as follows v = 2x + 1. Find, how
does x vary with time t. (Assume initial position to
be at x = 0)​

Answer 1

Given  : The velocity v of a moving particle changes with  displacement x as follows v = 2x + 1  

To Find : how  does x vary with time t.

Solution:

v  = 2x + 1

v  = dx/dt

=> dx/dt  =  2x  + 1

=> dx  = (2x+1)dt

=> dt  = dx/(2x + 1)  

Integrate both sides

=> ∫ dt   = ∫(1/(2x+1)) dx

=> t  = ln | 2x + 1| / 2   + C

at t = 0 x = 0

=> 0  = ln | 2(0) + 1| / 2   + C

=> 0  = ln | 1| / 2   + C

=> 0  =0 / 2   + C

=> C = 0

t  = ln | 2x + 1| / 2  

=> 2t =  ln | 2x + 1|

=> 2x + 1  = $e^{2t}$

=> x = ( $e^{2t}$  - 1)/2

This is how x varies with time t

Learn more:

Answer with full explaination!!!! - Brainly.in

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