Question

A particle is moving in a straight line and all the forces acting on it produce a constant power $P$. Calculate the distance travelled by the particle in the interval its speed increases frrom $v$ to $2v$.

• A. $\frac{m{v}^3}{3P}$
• B. $\frac{m{v}^3}{P}$
• C. $\frac{7m{v}^3}{3P}$
• D. $\frac{5m{v}^3}{3P}$

Answer 1

The correct option is C $\frac{7m{v}^3}{3P}$

Power P is constant

If sum of all forces is F & velocity

at any instant is V

then $P=FV$

write F = ma ; a= acceleration

or $F=mv\frac{dv}{dx};a=v\frac{dv}{dx}$ also.

So equation becomes $P=m\frac{vdv}{dx}.V$

it contains relation b/w v & x

$\Rightarrow \int \frac{p}{m}dx=\int v^{2}dv$

interactions both sides

taking limits $x=x_1$ to $x=x_2$

& $v=v$ to $v\to 2v$

we get

$\frac{p}{m}(x_2-x_1)=v^3{|}_v^{2v}=8v^3-v^3=7v^3$

$\Rightarrow$ distance $x_1-x_1=\frac{7v^3\times m}{p}$