1. A car is traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes and slows at 7.0 m/s². a) How far does the car go before it stops?

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**Physics Problem Example: Calculating Stopping Distance**

**Problem Statement:**
1. A car is traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes and slows at 7.0 m/s².

   a) How far does the car go before it stops?

**Detailed Breakdown:**

1. **Initial Conditions:**
   - Initial speed of the car (\(v_i\)): 20 m/s.
   - Reaction time of the driver: 0.80 s.

2. **Reaction Distance:**
   - During the reaction time, the car continues to travel at the initial speed. 
     Distance traveled during reaction time \(d_{reaction}\) can be calculated using the formula:
     \[
     d_{reaction} = v_i \times t_{reaction}
     \]
     Here,
     \[
     t_{reaction} = 0.80 \, \text{s}
     \]

     Calculation:
     \[
     d_{reaction} = 20 \, \text{m/s} \times 0.80 \, \text{s} = 16 \, \text{m}
     \]

3. **Braking Distance:**
   - After the brakes are applied, the car decelerates at \(7.0 \, \text{m/s}^2\). 
     We can use the kinematic equation to find the distance covered during braking (\(d_{braking}\)):

     \[
     v_f^2 = v_i^2 + 2 \times a \times d_{braking}
     \]

     Given:
     \[
     v_f = 0 \, \text{m/s} \quad (\text{since the car stops}), \quad a = -7.0 \, \text{m/s}^2
     \]

     Rearranging the equation to solve for \(d_{braking}\):
     \[
     0 = (20 \, \text{m/s})^2 + 2 \times (-7.0 \, \text{m/s}^2) \times d_{braking}
     \]

     \[
     0 = 400 \, \text{m}^2/\text{s}^2 - 14 \, \text{m/s
Transcribed Image Text:**Physics Problem Example: Calculating Stopping Distance** **Problem Statement:** 1. A car is traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes and slows at 7.0 m/s². a) How far does the car go before it stops? **Detailed Breakdown:** 1. **Initial Conditions:** - Initial speed of the car (\(v_i\)): 20 m/s. - Reaction time of the driver: 0.80 s. 2. **Reaction Distance:** - During the reaction time, the car continues to travel at the initial speed. Distance traveled during reaction time \(d_{reaction}\) can be calculated using the formula: \[ d_{reaction} = v_i \times t_{reaction} \] Here, \[ t_{reaction} = 0.80 \, \text{s} \] Calculation: \[ d_{reaction} = 20 \, \text{m/s} \times 0.80 \, \text{s} = 16 \, \text{m} \] 3. **Braking Distance:** - After the brakes are applied, the car decelerates at \(7.0 \, \text{m/s}^2\). We can use the kinematic equation to find the distance covered during braking (\(d_{braking}\)): \[ v_f^2 = v_i^2 + 2 \times a \times d_{braking} \] Given: \[ v_f = 0 \, \text{m/s} \quad (\text{since the car stops}), \quad a = -7.0 \, \text{m/s}^2 \] Rearranging the equation to solve for \(d_{braking}\): \[ 0 = (20 \, \text{m/s})^2 + 2 \times (-7.0 \, \text{m/s}^2) \times d_{braking} \] \[ 0 = 400 \, \text{m}^2/\text{s}^2 - 14 \, \text{m/s
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