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 } …