Question

The displacement x of a particle varies with time t as \(x=ae^{-  \alpha t} + be ^{\beta t}\), where \(a,b,\) \(\alpha\) and \(\beta\) are positive constants. The velocity of the particle will:

• A. go on decreasing with time
• B. be independent of \(\alpha\) and \(\beta\)
• C. drop to zero when \(\alpha\) = \(\beta\)
• D. go on increasing with time

Answer 1

The Correct Option is D

Given \(x=\)\(\,ae^{-\alpha t}+be^{\beta t}\)

Where \( a,b\),\(\,\alpha\), and \(\beta\) are positive constant 

\(V=\) \(\frac{dx}{dt}\) =\(\frac{d(ae^{-\alpha t}+be^{\beta t})}{dt}\) =\(−aαe^{ −αt}+ bβe^{βt }\)

∴ \(\frac{dx}{dt}\)=\(aα^2e^{-αt}+bβ^2e^{\beta t} \text{ is always >0}\) 

V is increasing the function of t.

Therefore, the correct option is (D): go on increasing with time.