Question

1. suppose that ( s(t) ) is the position in feet of a particle at time ( t ) seconds and that ( s(0)=6 mathrm{ft} ). suppose that ( s^{prime}(t) ), the velocity of the particle in ( mathrm{ft} / mathrm{s} ) at time ( t ) seconds, is shown on the graph.

a. find the displacement (change in position) from time ( t = 0 ) seconds to time ( t = 6 ) seconds.
b. find the total distance travelled from time ( t = 0 ) to ( t = 6 ).
c. find the position of the particle at time ( t = 4 ) seconds.

Explanation:

Step1: Recall the relationship between velocity and displacement

The displacement \( \Delta s\) from \(t = a\) to \(t = b\) is given by \( \Delta s=\int_{a}^{b}s^{\prime}(t)dt\), and the total - distance \(D\) is given by \(D = \int_{a}^{b}\vert s^{\prime}(t)\vert dt\). Also, \(s(t)=s(0)+\int_{0}^{t}s^{\prime}(u)du\).

Step2: Calculate the area of geometric shapes for \(s^{\prime}(t)\)

• For part A (displacement from \(t = 0\) to \(t = 6\)):

The area of a triangle is \(A=\frac{1}{2}bh\).
The area of the first triangle (from \(t = 0\) to \(t = 1\)): \(A_1=\frac{1}{2}(1)(4)=2\).
The area of the second triangle (from \(t = 1\) to \(t = 2.5\)): \(A_2=\frac{1}{2}(1.5)( - 4)=-3\).
The area of the third triangle (from \(t = 2.5\) to \(t = 6\)): The base of the third triangle from \(t = 2.5\) to \(t = 4\) is \(b_3 = 1.5\) and the height \(h_3\) at \(t = 4\) is \(y = 0\). Using the formula for the area of a trapezoid (or continue with triangle - like calculations). The slope of the line from \(t = 2.5\) to \(t = 6\): \(m=\frac{0-( - 4)}{4 - 2.5}=\frac{4}{1.5}=\frac{8}{3}\). The equation of the line \(y - (-4)=\frac{8}{3}(x - 2.5)\), \(y=\frac{8}{3}x-\frac{20}{3}-4=\frac{8}{3}x-\frac{32}{3}\). \(\int_{0}^{6}s^{\prime}(t)dt=\int_{0}^{1}s^{\prime}(t)dt+\int_{1}^{2.5}s^{\prime}(t)dt+\int_{2.5}^{6}s^{\prime}(t)dt\).
\(\int_{0}^{1}s^{\prime}(t)dt = 2\), \(\int_{1}^{2.5}s^{\prime}(t)dt=-3\), \(\int_{2.5}^{6}s^{\prime}(t)dt=\frac{1}{2}(6 - 2.5)\times\frac{8}{3}\times\frac{3}{8}\times(6 - 2.5)= \frac{1}{2}(3.5)\times\frac{8}{3}\times\frac{3}{8}\times(3.5)=3.5\).
\(\int_{0}^{6}s^{\prime}(t)dt=2-3 + 3.5=2.5\).

• For part B (total distance from \(t = 0\) to \(t = 6\)):

\(\vert\int_{0}^{1}s^{\prime}(t)dt\vert+\vert\int_{1}^{2.5}s^{\prime}(t)dt\vert+\vert\int_{2.5}^{6}s^{\prime}(t)dt\vert\).
\(\vert2\vert+\vert - 3\vert+\vert3.5\vert=2 + 3+3.5 = 8.5\).

• For part C (position at \(t = 4\)):

\(s(4)=s(0)+\int_{0}^{4}s^{\prime}(t)dt\).
\(\int_{0}^{4}s^{\prime}(t)dt=\int_{0}^{1}s^{\prime}(t)dt+\int_{1}^{2.5}s^{\prime}(t)dt+\int_{2.5}^{4}s^{\prime}(t)dt\).
\(\int_{0}^{1}s^{\prime}(t)dt = 2\), \(\int_{1}^{2.5}s^{\prime}(t)dt=-3\), \(\int_{2.5}^{4}s^{\prime}(t)dt=\frac{1}{2}(1.5)(4)=3\).
\(s(4)=6+(2-3 + 3)=8\).

Answer:

A. The displacement from \(t = 0\) to \(t = 6\) is \(2.5\) feet.
B. The total distance from \(t = 0\) to \(t = 6\) is \(8.5\) feet.
C. The position of the particle at \(t = 4\) is \(8\) feet.