Question

Consider a particle moving along the \(x\) -axis where \(x(t)\) is the position of the particle at time \(t, x^{\prime}(t)\) is its velocity, and \(\int_{a}^{b}\left|x^{\prime}(t)\right| d t\) is the distance the particle travels in the interval of time. A particle moves along the \(x\) -axis with velocity \(v(t)=1 / \sqrt{t}\) \(t > 0\). At time \(t=1,\) its position is \(x=4\). Find the total distance traveled by the particle on the interval \(1 \leq t \leq 4\).

Short answer

The total distance travelled by the particle in the interval \(1 \leq t \leq 4\) is 4 units.

Step-by-step solution

Evaluate the position function

Choose an antiderivative \(X(t)\) of \(v(t)\), so \(X(t) = \int v(t) dt = \int \frac{1} {\sqrt{t}} dt = 2\sqrt{t}\). Since at \(t=1\), \(x=4\), it follows that \(C=4-2(1)=2\). Therefore, the position function is \(X(t) = 2\sqrt{t} + 2\).

Calculate the total distance travelled

The total distance travelled by the particle is \(\int_{1}^{4} \left| v(t) \right| dt\). Given that \(v(t)\) is always positive for the given time interval, the integral simplifies to \(\int_{1}^{4} \frac{1} {\sqrt{t}} dt = \left[2\sqrt{t}\right]_{1}^{4} = 2\sqrt{4} - 2\sqrt{1} = 4\).