4. A biker is accelerating when she applies the brakes and then speeds up again. The net horizontal force is shown in the graph. Assume she moves in a straight line and the mass of the biker and bike is 80 kg. The biker was moving at 8 m/s at t = 0. How fast is she moving at t = 20 s? net x-Force (Newtons) 40 20 400 4 8 12 time (sec) 16 20

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Hi, I need help with number 4, thank you

### Problem Statement

A biker is accelerating when she applies the brakes and then speeds up again. The net horizontal force is shown in the graph. Assume she moves in a straight line and the mass of the biker and bike is 80 kg. The biker was moving at 8 m/s at \( t = 0 \). How fast is she moving at \( t = 20 \, \text{s} \)?

### Graph Description

The graph represents the net horizontal force applied on the biker over time. 

- On the y-axis, the net x-Force is measured in Newtons (N), with values ranging from -40 N to 40 N.
- On the x-axis, time is measured in seconds (sec), ranging from 0 to 20 seconds.

#### Graph Details:

1. From \( t = 0 \, \text{s} \) to \( t = 8 \, \text{s} \), the net x-Force is constant at 20 N.
2. At \( t = 8 \, \text{s} \), the net x-Force drops sharply to -20 N and remains constant until \( t = 12 \, \text{s} \).
3. From \( t = 12 \, \text{s} \) to \( t = 16 \, \text{s} \), the net x-Force increases linearly from -20 N to 10 N.
4. From \( t = 16 \, \text{s} \) to \( t = 20 \, \text{s} \), the net x-Force remains constant at 10 N.

### Solution Steps

1. **Calculation of Acceleration and Velocity:**
   - Use Newton's second law, \( F = ma \), to find acceleration for each time interval.
   
2. **First Interval (0 to 8 seconds):**
   - Force \( F = 20 \, \text{N} \)
   - Mass \( m = 80 \, \text{kg} \)
   - Acceleration \( a = \frac{F}{m} = \frac{20}{80} = 0.25 \, \text{m/s}^2 \)
   
   Velocity at \( t = 8 \, \text{s} \):
   \[ V_1 = V_0 + at = 8 + 0.25 \times
Transcribed Image Text:### Problem Statement A biker is accelerating when she applies the brakes and then speeds up again. The net horizontal force is shown in the graph. Assume she moves in a straight line and the mass of the biker and bike is 80 kg. The biker was moving at 8 m/s at \( t = 0 \). How fast is she moving at \( t = 20 \, \text{s} \)? ### Graph Description The graph represents the net horizontal force applied on the biker over time. - On the y-axis, the net x-Force is measured in Newtons (N), with values ranging from -40 N to 40 N. - On the x-axis, time is measured in seconds (sec), ranging from 0 to 20 seconds. #### Graph Details: 1. From \( t = 0 \, \text{s} \) to \( t = 8 \, \text{s} \), the net x-Force is constant at 20 N. 2. At \( t = 8 \, \text{s} \), the net x-Force drops sharply to -20 N and remains constant until \( t = 12 \, \text{s} \). 3. From \( t = 12 \, \text{s} \) to \( t = 16 \, \text{s} \), the net x-Force increases linearly from -20 N to 10 N. 4. From \( t = 16 \, \text{s} \) to \( t = 20 \, \text{s} \), the net x-Force remains constant at 10 N. ### Solution Steps 1. **Calculation of Acceleration and Velocity:** - Use Newton's second law, \( F = ma \), to find acceleration for each time interval. 2. **First Interval (0 to 8 seconds):** - Force \( F = 20 \, \text{N} \) - Mass \( m = 80 \, \text{kg} \) - Acceleration \( a = \frac{F}{m} = \frac{20}{80} = 0.25 \, \text{m/s}^2 \) Velocity at \( t = 8 \, \text{s} \): \[ V_1 = V_0 + at = 8 + 0.25 \times
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