A body of mass 2 kg, initially at rest on a smooth horizontal plane, is subjected to a horizontal force of magnitude $\frac{1}{2v + 1}$ N, where v is the velocity of the body (v > 0).

i) Find the time when the velocity is 1 $ms^{-1}$ .
ii) Find the displacement when the velocity is 1 $ms^{-1}$ .

Question

A body of mass 2 kg, initially at rest on a smooth horizontal plane, is subjected to a horizontal force of magnitude [latex]\frac{1}{2v + 1}[/latex] N, where v is the velocity of the body (v > 0).

i) Find the time when the velocity is 1[latex]ms^{-1}[/latex].
ii) Find the displacement when the velocity is 1[latex]ms^{-1}[/latex].

Accepted Answer

i) Using F = ma = m[latex]\frac{dv}{dt}[/latex]

[latex] \begin{matrix} \frac{1}{2v + 1} = 2 \frac{dv}{dt}. & \boxed{ ext{Write acceleration in } \frac{dv}{dt} ext{form since time is required.}} \end{matrix} [/latex]

Separating the variables gives

[latex] \int{dt} = \int{2(2v + 1)dv} [/latex]

⇒ t = 2v² + 2v + k.

When t = 0, v = 0 so k = 0 and therefore

t = 2v² + 2v.

When v = 1, t = 4. That is, when the velocity is 1[latex]ms^{-1}[/latex], the time is 4 seconds.

ii) F = ma = mv[latex]\frac{dv}{ds}[/latex]

[latex] \begin{matrix} \frac{1}{2v + 1} = 2 v\frac{dv}{ds}. & \boxed{ ext{Write acceleration in }v \frac{dv}{ds} ext{form since displacement is required.}} \end{matrix} [/latex]

Separating the variables gives

[latex] \int{ds} = \int{2v(2v + 1)dv} [/latex]

⇒ [latex]s = \frac{4}{3}v^{3} + v^{2} + k.[/latex]

When s = 0, v = 0 so k = 0 and therefore

[latex]s = \frac{4}{3}v^{3} + v^{2} [/latex].

When v = 1, [latex]s = \frac{7}{3}.[/latex] When the velocity is 1 [latex]ms^{-1}[/latex], the displacement is [latex]2\frac{1}{3}[/latex] m.

Question 18.3: A body of mass 2 kg, initially at rest on a smooth horizonta...

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