# Find relation between displacement and time

“A body is moving unidirectionally under the influence of a source of constant power. Find the relationship between displacement $$s$$ and time $$t$$.”

I proved it using dimensional homogeneity. Power has dimensions $${[ML^2T^{-3}]}.$$

As it is constant and mass M is also constant, So, $${[L^2T^{-3}]}$$is constant.

Hence $${L^2}\propto{T^3}$$

So, L $$\propto{T^{3/2}}$$

But this method feels like a trick proof or maybe not work in all situations (for example, here we had constant mass, but in other times we may not have that)

I feel that there is probably a better and concrete method through calculus. Can someone please help me out here.

Any help will be greatly appreciated.

• I find your solution much better and much more elegant that suggested in the answer. – sleepy May 23 at 10:57
• yes there is one method through calculus [commenting so I can get back to it and be forced to answer :)] – Adil Mohammed May 23 at 16:18

Is this the sort of thing you're looking for?

If the power is applied from time $$t=0$$, when the object is at rest, then at time $$t$$, $$\frac12 m v^2 =Pt.$$ Writing $$v=\frac{ds}{dt}$$, separating variables and preparing to integrate, $$\int_0^{s_1} ds =\int_0^{t_1} \sqrt{\frac{2Pt}{m}} dt$$ So $$s_1=\frac23 \sqrt{\frac{2P}{m}} t_1^{\ 3/2}$$ We can now drop the '1' subscript. It has served its purpose.

• Thanks, this is exactly what I was looking for! – InfiniteCool23 May 24 at 13:42
• I love your dimensions approach, though, as you know, it won't yield dimensionless numerical factors. – Philip Wood May 24 at 13:47
• Yes, that's why I wanted to know the calculus way, because I needed to put in real values and get the answer (which would have only been possible by knowig the proportionality constant) – InfiniteCool23 May 24 at 13:51

Your method works. You can also do this by considering that power $$P=\vec F \cdot \vec v = Fv$$ since the problem says "unidirectional" meaning $$\vec F$$ and $$\vec v$$ are in the same direction. Note also, that $$F=m\frac{v}{t}$$ so that $$P=m\frac{v^2}{t}$$ or $$Pt=mv^2$$ and noting that power and mass are constant, we can write $$v^2=kt$$ for some constant $$k$$. Remembering that $$v=\frac{x}{t}$$ we can also write $$\frac{x^2}{t^2} = kt$$ or $$x^2=kt^3$$ or $$x \propto t^\frac{3}{2}$$

• How did you write F=mv/t? – InfiniteCool23 May 23 at 6:21
• Acceleration a = v/t only when a is constant (w.r.t time) and initial velocity =0 – InfiniteCool23 May 23 at 6:21
• Because $F=ma$ and $a=v/t$. – joseph h May 23 at 6:22
• If the force is constant, then so is $a$. It also doesn't matter if the the initial velocity was not $0$. You get the exact same result. Try it. – joseph h May 23 at 6:24
• Force is not constant. Power is constant – InfiniteCool23 May 23 at 6:24