A 4.5 mile segment with an ADT of 42,376 vehicles per day experienced 43 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

   $$ADT \times 365 \times N \times L = 42376 \times 365 \times 5 \times 4.5 = 3.480e+8 veh-mi$$

3. Substituting

   $$R = (43 \times 10^{8})/3.480e+8 = 12.36 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$43(1 - 0.20) = 34.40 crashes expected$$

6. Prevented

   $$43 - 34.40 = 8.60 crashes over 5 years$$

Why the other options are there

• 9.56 (crashes per mile, no exposure)
• 43.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 4.5 mile segment with an ADT of 30,920 vehicles per day experienced 36 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

   $$ADT \times 365 \times N \times L = 30920 \times 365 \times 3 \times 4.5 = 1.524e+8 veh-mi$$

3. Substituting

   $$R = (36 \times 10^{8})/1.524e+8 = 23.63 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$36(1 - 0.30) = 25.20 crashes expected$$

6. Prevented

   $$36 - 25.20 = 10.80 crashes over 3 years$$

Why the other options are there

• 8.00 (crashes per mile, no exposure)
• 32.4 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 1.0 mile segment with an ADT of 14,427 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

   $$ADT \times 365 \times N \times L = 14427 \times 365 \times 3 \times 1.0 = 1.580e+7 veh-mi$$

3. Substituting

   $$R = (24 \times 10^{8})/1.580e+7 = 151.9 crashes per 100 MVM$$

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

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

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 1.5 mile segment with an ADT of 44,117 vehicles per day experienced 10 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

   $$ADT \times 365 \times N \times L = 44117 \times 365 \times 3 \times 1.5 = 7.246e+7 veh-mi$$

3. Substituting

   $$R = (10 \times 10^{8})/7.246e+7 = 13.80 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$10(1 - 0.35) = 6.50 crashes expected$$

6. Prevented

   $$10 - 6.50 = 3.50 crashes over 3 years$$

Why the other options are there

• 6.67 (crashes per mile, no exposure)
• 10.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 5.5 mile segment with an ADT of 10,246 vehicles per day experienced 11 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

   $$ADT \times 365 \times N \times L = 10246 \times 365 \times 3 \times 5.5 = 6.171e+7 veh-mi$$

3. Substituting

   $$R = (11 \times 10^{8})/6.171e+7 = 17.83 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$11(1 - 0.30) = 7.70 crashes expected$$

6. Prevented

   $$11 - 7.70 = 3.30 crashes over 3 years$$

Why the other options are there

• 2.00 (crashes per mile, no exposure)
• 9.9 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 5.0 mile segment with an ADT of 44,599 vehicles per day experienced 33 crashes in 5 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

   $$ADT \times 365 \times N \times L = 44599 \times 365 \times 5 \times 5.0 = 4.070e+8 veh-mi$$

3. Substituting

   $$R = (33 \times 10^{8})/4.070e+8 = 8.11 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$33(1 - 0.25) = 24.75 crashes expected$$

6. Prevented

   $$33 - 24.75 = 8.25 crashes over 5 years$$

Why the other options are there

• 6.60 (crashes per mile, no exposure)
• 41.3 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 4.0 mile segment with an ADT of 10,600 vehicles per day experienced 41 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

   $$ADT \times 365 \times N \times L = 10600 \times 365 \times 3 \times 4.0 = 4.643e+7 veh-mi$$

3. Substituting

   $$R = (41 \times 10^{8})/4.643e+7 = 88.31 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$41(1 - 0.30) = 28.70 crashes expected$$

6. Prevented

   $$41 - 28.70 = 12.30 crashes over 3 years$$

Why the other options are there

• 10.25 (crashes per mile, no exposure)
• 36.9 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 2.0 mile segment with an ADT of 23,166 vehicles per day experienced 49 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

   $$ADT \times 365 \times N \times L = 23166 \times 365 \times 5 \times 2.0 = 8.456e+7 veh-mi$$

3. Substituting

   $$R = (49 \times 10^{8})/8.456e+7 = 57.95 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$49(1 - 0.20) = 39.20 crashes expected$$

6. Prevented

   $$49 - 39.20 = 9.80 crashes over 5 years$$

Why the other options are there

• 24.50 (crashes per mile, no exposure)
• 49.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 3.5 mile segment with an ADT of 44,762 vehicles per day experienced 22 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

   $$ADT \times 365 \times N \times L = 44762 \times 365 \times 3 \times 3.5 = 1.716e+8 veh-mi$$

3. Substituting

   $$R = (22 \times 10^{8})/1.716e+8 = 12.82 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$22(1 - 0.35) = 14.30 crashes expected$$

6. Prevented

   $$22 - 14.30 = 7.70 crashes over 3 years$$

Why the other options are there

• 6.29 (crashes per mile, no exposure)
• 23.1 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction

A 2.0 mile segment with an ADT of 8,662 vehicles per day experienced 11 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

   $$ADT \times 365 \times N \times L = 8662 \times 365 \times 5 \times 2.0 = 3.162e+7 veh-mi$$

3. Substituting

   $$R = (11 \times 10^{8})/3.162e+7 = 34.79 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$11(1 - 0.30) = 7.70 crashes expected$$

6. Prevented

   $$11 - 7.70 = 3.30 crashes over 5 years$$

Why the other options are there

• 5.50 (crashes per mile, no exposure)
• 16.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Reduction