Practice portal › Laws of Motion › Newton's Laws and Free-Body Diagrams
Asked in JEE Main Online 2016 · Position- and velocity-dependent forces
Idea: integrate the law, then look for the pair of variables that leaves a straight line.
m(dv)/(dt)=R/(t²)v, and separating the variables,
(dv)/v=R/m(dt)/(t²).
Integrating both sides,
ln v=-R/m·1/t+C.
That is of the form y=mx+c with y=log v and x=1/t, so those are the axes that give a straight line — and its slope then measures R/m.
The minus sign in front means the line slopes downward, which is also a useful check on the data.
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