Determine the time interval and displacement of a particle when speed changes from 1 m/s to 3 m/s.

$$a=100-4v^2 (in m/s^2) $$

$ i.e\space \space\dfrac { dv}{dt}=100-4v^2 ----------(i) \\ dt=\dfrac {dv}{100-4v^2 }$

Time taken to increase the velocity from 1m/s to 3 m/s is found out by integrating the above equation from v=1m/s to v=3m/s

$ t=∫_1^3 \dfrac 1{100-4v^2 } dv \\ =\dfrac {-1}4 ∫_1^3 \dfrac 1{v^2-5^2 } \\ =\dfrac {-1}2.\dfrac1{10} ln⁡[\dfrac {5-v}{5+v}]_(v=1)^(v=3) \\ =\dfrac {-1}{20}[ln⁡(\dfrac28)-ln⁡(\dfrac46) \\ \therefore t= 0.245s $

We know that,

$ v = \dfrac {dx}{dt} \text { and } a=\dfrac {dv}{dt} \\ \therefore \dfrac va= \dfrac {dx}{dv} \\ dx=\dfrac {vdv}a= \dfrac{vdv}{100-4v^2} $

$\therefore $ Displacement when velocity changes from 1m/s to 3m/s

$ x=∫_1^3 \dfrac 1{100-4v^2} dv \\ =\dfrac {-1}4 ∫_1^3 \dfrac {\frac d {dv} v^2-25}{v^2-25}.\dfrac12 dv \\ =\dfrac {-1}4 [ln⁡|v^2-25|]_1^3 \\ x=0.051m \Rightarrow Ans$