2023TrialShort answer

Q15 (a) (ii)

3 marks

Question

A particle of unit mass is moving in horizontal motion, subject to a resistance force of $v^2+v^3$ , where $v$ is the object’s velocity. The particle has initial velocity $v_0$ , where $v_0>0$ .

Show that the time $T$ taken to travel the distance $s$ is

T=\frac{1}{v_0}-s.

Try the question before revealing the answer.

Solution

x=\ln\left(\frac{v_0(1+v)}{v(1+v_0)}\right)

\frac{v_0(1+v)}{v(1+v_0)}=e^x

\frac{1+v}{v}=e^x\left(\frac{1+v_0}{v_0}\right)

1+\frac1v=e^x\left(\frac{1+v_0}{v_0}\right)

\frac1v=e^x\left(\frac{1+v_0}{v_0}\right)-1

\frac{dt}{dx}=e^x\left(\frac{1+v_0}{v_0}\right)-1

\int_0^Tdt=\int_0^s\left[e^x\left(\frac{1+v_0}{v_0}\right)-1\right]dx

T=\left[e^x\left(\frac{1+v_0}{v_0}\right)-x\right]_0^s

=\left[e^s\left(\frac{1+v_0}{v_0}\right)-s\right]-\left(\frac{1+v_0}{v_0}\right)

=\left(\frac{v_0+2}{v_0+1}\right)\left(\frac{1+v_0}{v_0}\right)-s-\left(\frac{1+v_0}{v_0}\right)

=\left(\frac{v_0+2}{v_0}\right)-\left(\frac{1+v_0}{v_0}\right)-s

=\frac1{v_0}-s

Alternate solution:

\frac{dv}{dt}=-(v^2+v^3)

=-v^2(1+v)

\int_{v_0}^{v}\frac{dv}{v^2(1+v)}=-\int_0^t dt

-t=\int_{v_0}^{v}\left(\frac1{v+1}-\frac1v+\frac1{v^2}\right)dv

=\int_{v_0}^{v}\left(\frac1{v+1}-\frac1v\right)dv+\int_{v_0}^{v}\frac1{v^2}dv

=x+\left[-\frac1v\right]_{v_0}^{v}\qquad\text{(from part i.)}

=x+\frac1{v_0}-\frac1v

t=-x-\frac1{v_0}+\frac1v

t=T,\quad v=\frac{v_0}{2},\quad x=s

T=-s-\frac1{v_0}+\frac2{v_0}

=\frac1{v_0}-s

ii.

\frac{1}{v^2(v-1)}\equiv\frac{av+b}{v^2}+\frac{c}{v-1}

(av+b)(v-1)+cv^2\equiv1

v=1\Rightarrow c=1

v=0\Rightarrow b=-1

Comparing coefficients of $v^2\Rightarrow(a+c)=0$ .

\therefore a=-1

\therefore\frac{1}{v^2(v-1)}\equiv\frac1{v-1}-\frac1v-\frac1{v^2}

\frac{dv}{dt}=-v^2+v^3

\int_0^Tdt=\int_{v_0}^{v_0/2}\left(\frac1{v-1}-\frac1v-\frac1{v^2}\right)dv

T=\left[\ln\left|\frac{v-1}{v}\right|+\frac1v\right]_{v_0}^{v_0/2}

\begin{aligned} &= \ln\left|\frac{v_0\left(\frac{v_0}{2}-1\right)}{\frac{v_0}{2}(v_0-1)}\right|+\frac{1}{v_0/2}-\frac{1}{v_0}\\ &=\ln\left|\frac{v_0-2}{v_0-1}\right|+\frac{1}{v_0}\\ &=s+\frac{1}{v_0} \end{aligned}

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Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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