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Question

The position function of a particle is given by $$x\left(t\right)=k{t}^{{5}/{2}}$$, where $$k$$ is a constant.
If the particle starts at rest and is propelled through some distance $$d$$ so that the trajectory matches $$x\left(t\right)$$, the work done on the particle is proportional to which power of $$t$$?


A
t5
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B
t3
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C
t5/2
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D
t3/2
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E
Not enough information
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Solution

The correct option is A $${t}^{3}$$
$$x=kt^{5/2}$$
Speed of the particle is $$v=\dfrac{dx}{dt}=\dfrac{5}{2}kt^{3/2}$$

The acceleration of the particle becomes $$a=\dfrac{dv}{dt}=\dfrac{15}{4}kt^{1/2}$$

Force acting on particle is given by $$ma=\dfrac{15}{4}mkt^{1/2}$$

Hence the work done is given by $$F.x\propto t^{3}$$

Physics

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