Kinematics-1D

The given position at $t = 1.0\text{ s}$ is $x(1.0) = 2.0\text{ m}$. We need to find $x(6.0)$, which is given by:
$x(6.0) = x(1.0) + \Delta x_{1 \to 6}$
From the graph, we can break the interval $t = 1.0\text{ s}$ to $t = 6.0\text{ s}$ into two sections:
1. From $t = 1.0\text{ s}$ to $t = 2.0\text{ s}$:
The region forms a triangle above the time axis with base $1.0\text{ s}$ and height $2.0\text{ m/s}$.
$\text{Area}_1 = \frac{1}{2} \times 1.0 \times 2.0 = +1.0\text{ m}$
2. From $t = 2.0\text{ s}$ to $t = 6.0\text{ s}$:
The region forms a triangle below the time axis with base $4.0\text{ s}$ (from $t = 2.0$ to $t = 6.0$) and height $-2.0\text{ m/s}$.
$\text{Area}_2 = \frac{1}{2} \times 4.0 \times (-2.0) = -4.0\text{ m}$
Total displacement from $t = 1.0\text{ s}$ to $t = 6.0\text{ s}$:
$\Delta x_{1 \to 6} = \text{Area}_1 + \text{Area}_2 = 1.0\text{ m} - 4.0\text{ m} = -3.0\text{ m}$
Calculating the position at $t = 6.0\text{ s}$:
$x(6.0) = x(1.0) + \Delta x_{1 \to 6} = 2.0\text{ m} + (-3.0\text{ m}) = -1.0\text{ m}$

Given data:
Initial speed $v_i = 0\text{ m/s}$ (since the skier starts from rest)
Final speed $v_f = 8.0\text{ m/s}$
Time taken $\Delta t = 5.0\text{ s}$
Using the formula for average acceleration:
$a_{\text{avg}} = \frac{8.0\text{ m/s} - 0\text{ m/s}}{5.0\text{ s}} = \frac{8.0}{5.0}\text{ m/s}^2 = 1.6\text{ m/s}^2$