Let's solve each activity one by one:
### Activity 1:
- **Given:**
- Acceleration, `a` = 3 m/s²
- Time, `t` = 5 seconds
- Initial velocity, `u` = 0 m/s
We can use the kinematic equation for uniformly accelerated motion to find the distance `s` traveled:
\[ s = ut + \frac{1}{2}at^2 \]
Let's plug in the values:
\[ s = (0 \, m/s \cdot 5 \, s) + \frac{1}{2} \cdot (3 \, m/s² \cdot (5 \, s)^2) \]
\[ s = 0 + \frac{1}{2} \cdot 3 \cdot 25 \]
\[ s = \frac{3}{2} \cdot 25 \]
\[ s = 1.5 \cdot 25 \]
\[ s = 37.5 \, m \]
The skateboarder has traveled 37.5 meters.
### Activity 2:
- **Given:**
- Initial velocity, `u` = 6 m/s
- Acceleration, `a` = 2 m/s²
- Distance, `s` = 75 m
We can use another kinematic equation to find the final velocity `v`:
\[ v^2 = u^2 + 2as \]
Substitute the given values:
\[ v^2 = (6 \, m/s)^2 + 2 \cdot (2 \, m/s² \cdot 75 \, m) \]
\[ v^2 = 36 + 2 \cdot 150 \]
\[ v^2 = 36 + 300 \]
\[ v^2 = 336 \]
Take the square root of both sides to find `v`:
\[ v = \sqrt{336} \]
\[ v \approx 18.33 \, m/s \]
The final velocity of the object is approximately 18.33 m/s.
### Activity 3:
- **Given:**
- Initial velocity, `u` = 20 m/s
- Acceleration, `a` = 8 m/s²
- Time, `t` = 10 seconds
Using the same kinematic equation as in Activity 1 to find position `s`:
\[ s = ut + \frac{1}{2}at^2 \]
Substitute the values:
\[ s = (20 \, m/s \cdot 10 \, s) + \frac{1}{2} \cdot (8 \, m/s² \cdot (10 \, s)^2) \]
\[ s = 200 + \frac{1}{2} \cdot 8 \cdot 100 \]
\[ s = 200 + 4 \cdot 100 \]
\[ s = 200 + 400 \]
\[ s = 600 \, m \]
The position of the car at the end of the 10 seconds is 600 meters from the starting point.
