Hello,

I am stuck on this mechanics question in the link below. Can anyone please help me out? Thanks

\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]

I can't see why \(c^2\) should equal 1 here - not sure where your explanation comes from. The next step here is to do what you would normally do in 2U, i.e. sub in the boundaries.

\begin{align*} \frac1{2k} \left[ \ln (c^2) - \ln (c^2+c^2) \right] &= -x\\ \frac{1}{2k} \ln \left(\frac{c^2}{c^2+c^2}\right)&=-x \tag{log laws}\\ \frac{1}{2k}\ln \frac12 &=-x\\ \therefore x &= -\frac1{2k}\ln \frac12\\ &= \frac1{2k}\ln 2. \end{align*}

\[ \text{Also I think there's a mistake with the velocity.}\\ \text{From }\int -dt = \frac1k\int \frac{dv}{c^2+v^2}\text{ you should've obtained }\boxed{-t = \frac{1}{k} \left( \frac{1}{c}\tan^{-1} \frac{v}{c}\right)+C} .\]

Note that whilst it is true that \( \int \frac{x}{a^2+x^2}dx = \frac12 \ln (a^2+x^2)+C \), when the numerator is instead a constant you have \( \int \frac{1}{a^2+x^2}\,dx = \frac1a \tan^{-1} \frac{x}a \). The first is a \( \int \frac{f^\prime(x)}{f(x)}\,dx \) pattern, but the second is a 3U standard integral from the inverse trig topic.