Exercise 3

We treat the expressions from Exercise 1 as acceleration for a particle. Hence

\[a(t) = \ddot{x}(t) = t^4 \] \[v(t) = \dot{x}(t) = \int t^4 \, \mathrm{d}t = \frac{1}{5} t^5 + v_0\]

The constant of integration \(C\) becomes the initial velocity \(v_0 = v(t=0)\). The value for the initial velocity is not given and cannot be derived solely from the acceleration.

In the task, the velocity is specified as a definite integral with a fixed lower integration limit. The initial velocity can be derived from this lower integration limit:

\[v(t) = \int_0^t t'^4 \mathrm{d}t' \Rightarrow v(0) = \int_0^0 t'^4 \mathrm{d}t' = 0 \] \[ \Rightarrow v(t) = \frac{1}{5} t^5 \]

The function for the position is given by: \[x(t) = \int v(t) \, \mathrm{d}t = \int \frac{1}{5} t^5 \, \mathrm{d}t = \frac{1}{30} t^6 + x_0\]

\(x_0 = x(t=0)\) denotes the position of the particle at time \(t=0\) and is a constant of integration, which can be chosen freely.

If you want \(x_0\) to be zero, this can be achived by \[ x(t) = \int_0^t v(t') \mathrm{d}t' \Rightarrow x(0) = \int_0^0 v(t') \mathrm{d}t' = 0 \]

The intervalls of integration for velocity and position need not be identical! The initial value for velocity and position are given by the lower integration limits. It is not mandatory that both values are zero.

Similar considerations also apply to the other two expressions.

---

\[\dot{x}(t) = \int_0^t \cos t'\, \mathrm{d}t' = \sin t \] \[ \Rightarrow v_0 = 0 \] \[x(t) = \int_0^t ( \sin t' )\, \mathrm{d}t' = [-\cos t']_0^t = -\cos t - ( - \cos 0) = - \cos t + 1 \] \[ \Rightarrow x_0 = 0 \]

---

\[\dot{x}(t) = \int_0^t (t'^2-2)\,\mathrm{d}t' = \frac{1}{3} t^3 - 2 t \] \[ \Rightarrow v_0 = 0 \] \[x(t) = \int_0^t \left( \frac{1}{3} t'^3 - 2 t'\right) \mathrm{d}t' = \frac{1}{12} t^4 - t^2 \] \[ \Rightarrow x_0 = 0 \]