A 2.0 mile segment with an ADT of 31,354 vehicles per day experienced 20 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 25% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 31354 \times 365 \times 3 \times 2.0 = 6.867e+7 veh-mi$$

3. Substituting

   $$R = (20 \times 10^{8})/6.867e+7 = 29.13 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$20(1 - 0.25) = 15.00 crashes expected$$

6. Prevented

   $$20 - 15.00 = 5.00 crashes over 3 years$$

Why the other options are there

• 10.00 (crashes per mile, no exposure)
• 15.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 5.0 mile segment with an ADT of 15,775 vehicles per day experienced 26 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 40% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 15775 \times 365 \times 5 \times 5.0 = 1.439e+8 veh-mi$$

3. Substituting

   $$R = (26 \times 10^{8})/1.439e+8 = 18.06 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$26(1 - 0.40) = 15.60 crashes expected$$

6. Prevented

   $$26 - 15.60 = 10.40 crashes over 5 years$$

Why the other options are there

• 5.20 (crashes per mile, no exposure)
• 52.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 4.5 mile segment with an ADT of 35,816 vehicles per day experienced 24 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 30% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 35816 \times 365 \times 3 \times 4.5 = 1.765e+8 veh-mi$$

3. Substituting

   $$R = (24 \times 10^{8})/1.765e+8 = 13.60 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$24(1 - 0.30) = 16.80 crashes expected$$

6. Prevented

   $$24 - 16.80 = 7.20 crashes over 3 years$$

Why the other options are there

• 5.33 (crashes per mile, no exposure)
• 21.6 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 1.5 mile segment with an ADT of 9,555 vehicles per day experienced 40 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 25% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 9555 \times 365 \times 3 \times 1.5 = 1.569e+7 veh-mi$$

3. Substituting

   $$R = (40 \times 10^{8})/1.569e+7 = 254.9 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$40(1 - 0.25) = 30.00 crashes expected$$

6. Prevented

   $$40 - 30.00 = 10.00 crashes over 3 years$$

Why the other options are there

• 26.67 (crashes per mile, no exposure)
• 30.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 5.5 mile segment with an ADT of 31,028 vehicles per day experienced 35 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 20% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 31028 \times 365 \times 5 \times 5.5 = 3.114e+8 veh-mi$$

3. Substituting

   $$R = (35 \times 10^{8})/3.114e+8 = 11.24 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$35(1 - 0.20) = 28.00 crashes expected$$

6. Prevented

   $$35 - 28.00 = 7.00 crashes over 5 years$$

Why the other options are there

• 6.36 (crashes per mile, no exposure)
• 35.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 2.0 mile segment with an ADT of 37,491 vehicles per day experienced 58 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 35% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 37491 \times 365 \times 5 \times 2.0 = 1.368e+8 veh-mi$$

3. Substituting

   $$R = (58 \times 10^{8})/1.368e+8 = 42.38 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$58(1 - 0.35) = 37.70 crashes expected$$

6. Prevented

   $$58 - 37.70 = 20.30 crashes over 5 years$$

Why the other options are there

• 29.00 (crashes per mile, no exposure)
• 101.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 3.5 mile segment with an ADT of 33,511 vehicles per day experienced 59 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 45% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 33511 \times 365 \times 3 \times 3.5 = 1.284e+8 veh-mi$$

3. Substituting

   $$R = (59 \times 10^{8})/1.284e+8 = 45.94 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$59(1 - 0.45) = 32.45 crashes expected$$

6. Prevented

   $$59 - 32.45 = 26.55 crashes over 3 years$$

Why the other options are there

• 16.86 (crashes per mile, no exposure)
• 79.7 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 3.5 mile segment with an ADT of 36,887 vehicles per day experienced 21 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 35% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 36887 \times 365 \times 5 \times 3.5 = 2.356e+8 veh-mi$$

3. Substituting

   $$R = (21 \times 10^{8})/2.356e+8 = 8.91 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$21(1 - 0.35) = 13.65 crashes expected$$

6. Prevented

   $$21 - 13.65 = 7.35 crashes over 5 years$$

Why the other options are there

• 6.00 (crashes per mile, no exposure)
• 36.8 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 5.5 mile segment with an ADT of 37,570 vehicles per day experienced 56 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 30% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 37570 \times 365 \times 5 \times 5.5 = 3.771e+8 veh-mi$$

3. Substituting

   $$R = (56 \times 10^{8})/3.771e+8 = 14.85 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$56(1 - 0.30) = 39.20 crashes expected$$

6. Prevented

   $$56 - 39.20 = 16.80 crashes over 5 years$$

Why the other options are there

• 10.18 (crashes per mile, no exposure)
• 84.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

A 3.0 mile segment with an ADT of 17,701 vehicles per day experienced 48 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 35% crash reduction factor.

Start with the thinking

• Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
• A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

1. Formula

   $$R = (A \times 100,000,000)/(ADT \times 365 \times N \times L)$$

2. Exposure

   $$ADT \times 365 \times N \times L = 17701 \times 365 \times 3 \times 3.0 = 5.815e+7 veh-mi$$

3. Substituting

   $$R = (48 \times 10^{8})/5.815e+7 = 82.55 crashes per 100 MVM$$

4. Formula

   $$crashes after = crashes \times (1 - CRF)$$

5. Substituting

   $$48(1 - 0.35) = 31.20 crashes expected$$

6. Prevented

   $$48 - 31.20 = 16.80 crashes over 3 years$$

Why the other options are there

• 16.00 (crashes per mile, no exposure)
• 50.4 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments