Answer:
Between time interval 20 s te 40 s, there is non-zero acceleration and retardation. Hence, distance travelled during this interval
=
Area between time interval 20 s to 40 s
=
1
2
×
20
×
3
+
20
×
1
=
30
+
20
=
50
m
Explanation:
Explanation:
It seems you are referring to a graph that provides the velocity of an object over time, but since the graph is not visible to me, I'll guide you through how to approach solving these problems using typical methods based on velocity-time graphs.
### a) Determining the Distance Traveled Between \( t = 2 \)s and \( t = 10 \)s
To find the distance traveled, you need to calculate the area under the velocity-time graph between \( t = 2 \) seconds and \( t = 10 \) seconds.
1. **Identify the shape(s) under the graph between these times**:
- If the graph forms simple geometric shapes like rectangles, triangles, or trapezoids, calculate the area of these shapes.
- Sum these areas to get the total distance.
2. **Example Calculation**:
- Suppose the graph forms a trapezoid from \( t = 2 \) to \( t = 4 \) and a rectangle from \( t = 4 \) to \( t = 10 \).
- For the trapezoid, the area is given by \(\frac{1}{2} \times (b_1 + b_2) \times h\), where \( b_1 \) and \( b_2 \) are the parallel sides (velocities at \( t = 2 \) and \( t = 4 \)), and \( h \) is the time interval.
- For the rectangle, the area is given by \( \text{length} \times \text{width} \).
### b) Finding the Maximum Retardation (Deceleration)
Retardation (or deceleration) is the negative acceleration. The maximum retardation can be found by looking at the steepest negative slope on the velocity-time graph.
1. **Identify the steepest negative slope**:
- The slope of a velocity-time graph at any point gives the acceleration.
- Calculate the slope (change in velocity divided by the change in time) for the steepest negative slope section of the graph.
2. **Example Calculation**:
- Suppose the graph drops from 4 m/s to 0 m/s between \( t = 6 \) and \( t = 8 \).
- The slope (retardation) = \(\frac{0 - 4}{8 - 6} = \frac{-4}{2} = -2 \) m/s².
### c) Calculating the Acceleration of the Object
Acceleration is the rate of change of velocity with respect to time. To find the acceleration at different points or intervals on the graph, follow these steps:
1. **Identify the slope at different points**:
- For intervals where the velocity is changing, calculate the slope as \(\frac{\Delta v}{\Delta t}\).
2. **Example Calculation**:
- If the velocity increases from 0 m/s to 4 m/s between \( t = 0 \) and \( t = 2 \), the acceleration is \(\frac{4 - 0}{2 - 0} = 2 \) m/s².
To provide more precise answers, I would need specific points or coordinates from the graph. If you can describe the graph in more detail or provide key coordinates (like velocity values at specific times), I can give more specific guidance on the calculations.
