A) Proportional to t
B) Proportional to s
C) s
D) Constant
Correct Answer: C
Solution :
Given that \[s=a{{e}^{t}}+\frac{b}{{{e}^{t}}}\] Differentiating w.r.t. time t, we get \[\frac{ds}{dt}=\]velocity \[=a{{e}^{t}}-\frac{b}{{{e}^{t}}}\] Again \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\] acceleration \[=a{{e}^{t}}+\frac{b}{{{e}^{t}}}=s\].You need to login to perform this action.
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