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. $t^5$
• B. $t^3$
• C. $t^{5/2}$
• D. $t^{3/2}$
• E. Not enough information

Answer 1

Answer: (B)

$t^3$

$x=k{t}^{5/2}$

Speed of the particle is $v=\frac{dx}{dt}=\frac{5}{2}k{t}^{3/2}$

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

Force acting on particle is given by $ma=\frac{15}{4}mk{t}^{1/2}$

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