Answer: \(k=10.0\), approximately.

Use the given velocity, acceleration or displacement relationship consistently. Remember that acceleration is the derivative of velocity, and displacement is obtained by integrating velocity.

The total distance is the integral of the velocity over the two time intervals:

\(\int_0^4 e^{t/4}\,dt+\int_4^k\frac{16e}{t^2}\,dt=13.4.\)

For the first integral,

\(\int e^{t/4}\,dt=4e^{t/4}.\)

So

\(\int_0^4 e^{t/4}\,dt =\left[4e^{t/4}\right]_0^4 =4e-4.\)

For the second integral,

\(\int\frac{16e}{t^2}\,dt =16e\int t^{-2}\,dt =-\frac{16e}{t}.\)

Thus

\(\int_4^k\frac{16e}{t^2}\,dt =\left[-\frac{16e}{t}\right]_4^k =-\frac{16e}{k}+4e.\)

Therefore

\(4e-4+4e-\frac{16e}{k}=13.4.\)

So

\(8e-4-\frac{16e}{k}=13.4.\)

Rearranging,

\(\frac{16e}{k}=8e-17.4.\)

Hence

\(k=\frac{16e}{8e-17.4}.\)

Therefore

\(k=10.0\)

to 3 significant figures.