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Question

Physics Question on Motion in a straight line

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

A

be independent of β\beta

B

drop to zero when α=β\alpha = \beta

C

go on decreasing with time

D

go on increasing with time

Answer

go on increasing with time

Explanation

Solution

x=aeαt+beβtx = ae^{-\alpha t}+be^{\beta t}
dxdt=aαeαt+bβeβt\frac{dx}{dt} = -a \alpha e^{-\alpha t}+b \beta e^{\beta t}
v=aαeαt+bβeβtv = -a \alpha e^{-\alpha t}+b \beta e^{\beta t}
For certain value of t, velocity will increases.