0
$\begingroup$

In common examples and understanding, like bullet firing, higher the velocity the distance traveled will be higher in a given time, but in the case of terminal velocity concept it is said that the drag force is proportional to the velocity (in the stokes law). If so, the object with higher velocity will have difficulty to travel a certain distance compared to object with lower velocity.

It seems contradictory to the common understanding.

What are the things I am missing to understand that law?

Additional doubt, is stokes law confined to only spherical objects?

$\endgroup$

1 Answer 1

0
$\begingroup$

If the initial velocity is $v_0$ and $\frac{dv}{dt}=-k(v-v_T)$ (where $v_T$ is the terminal velocity), then $$v=\frac{dx}{dt}=v_T+(v_0-v_T)e^{-kt}$$ So $$ x=v_Tt+\frac{(v_0-v_T)}{k}\left(1-e^{-kt}\right)$$At very long times, $$x=v_Tt+\frac{(v_0-v_T)}{k}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.