A) proportional to\[{{s}^{2}}\]
B) proportional to \[\frac{1}{{{s}^{2}}}\].
C) proportional to \[s\].
D) proportional to\[\frac{1}{s}\].
E) a constant
Correct Answer: E
Solution :
According to question, \[s\propto {{v}^{2}}\] \[\Rightarrow \] \[v=k\sqrt{s}\] Now, acceleration \[=\frac{dv}{dt}=k\frac{1}{2\sqrt{s}}.\frac{ds}{dt}\] \[=k\frac{1}{2\sqrt{s}}.k\sqrt{s}\] \[=\frac{{{k}^{2}}}{2}=a\,constant\]
You need to login to perform this action.
You will be redirected in
3 sec
