A 5.5 mile segment with an ADT of 16,036 vehicles per day experienced 51 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

   $$ADT \times 365 \times N \times L = 16036 \times 365 \times 5 \times 5.5 = 1.610e+8 veh-mi$$

3. Substituting

   $$R = (51 \times 10^{8})/1.610e+8 = 31.68 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$51(1 - 0.40) = 30.60 crashes expected$$

6. Prevented

   $$51 - 30.60 = 20.40 crashes over 5 years$$

Why the other options are there

• 9.27 (crashes per mile, no exposure)
• 102.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 4.5 mile segment with an ADT of 39,911 vehicles per day experienced 8 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

   $$ADT \times 365 \times N \times L = 39911 \times 365 \times 5 \times 4.5 = 3.278e+8 veh-mi$$

3. Substituting

   $$R = (8 \times 10^{8})/3.278e+8 = 2.44 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$8(1 - 0.25) = 6.00 crashes expected$$

6. Prevented

   $$8 - 6.00 = 2.00 crashes over 5 years$$

Why the other options are there

• 1.78 (crashes per mile, no exposure)
• 10.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 5.5 mile segment with an ADT of 11,507 vehicles per day experienced 44 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

   $$ADT \times 365 \times N \times L = 11507 \times 365 \times 3 \times 5.5 = 6.930e+7 veh-mi$$

3. Substituting

   $$R = (44 \times 10^{8})/6.930e+7 = 63.49 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$44(1 - 0.35) = 28.60 crashes expected$$

6. Prevented

   $$44 - 28.60 = 15.40 crashes over 3 years$$

Why the other options are there

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

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 4.5 mile segment with an ADT of 43,731 vehicles per day experienced 17 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

   $$ADT \times 365 \times N \times L = 43731 \times 365 \times 5 \times 4.5 = 3.591e+8 veh-mi$$

3. Substituting

   $$R = (17 \times 10^{8})/3.591e+8 = 4.73 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$17(1 - 0.30) = 11.90 crashes expected$$

6. Prevented

   $$17 - 11.90 = 5.10 crashes over 5 years$$

Why the other options are there

• 3.78 (crashes per mile, no exposure)
• 25.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 5.0 mile segment with an ADT of 14,740 vehicles per day experienced 18 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

   $$ADT \times 365 \times N \times L = 14740 \times 365 \times 5 \times 5.0 = 1.345e+8 veh-mi$$

3. Substituting

   $$R = (18 \times 10^{8})/1.345e+8 = 13.38 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$18(1 - 0.25) = 13.50 crashes expected$$

6. Prevented

   $$18 - 13.50 = 4.50 crashes over 5 years$$

Why the other options are there

• 3.60 (crashes per mile, no exposure)
• 22.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 5.0 mile segment with an ADT of 42,828 vehicles per day experienced 43 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

   $$ADT \times 365 \times N \times L = 42828 \times 365 \times 3 \times 5.0 = 2.345e+8 veh-mi$$

3. Substituting

   $$R = (43 \times 10^{8})/2.345e+8 = 18.34 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$43(1 - 0.25) = 32.25 crashes expected$$

6. Prevented

   $$43 - 32.25 = 10.75 crashes over 3 years$$

Why the other options are there

• 8.60 (crashes per mile, no exposure)
• 32.3 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 4.5 mile segment with an ADT of 26,703 vehicles per day experienced 19 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

   $$ADT \times 365 \times N \times L = 26703 \times 365 \times 3 \times 4.5 = 1.316e+8 veh-mi$$

3. Substituting

   $$R = (19 \times 10^{8})/1.316e+8 = 14.44 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$19(1 - 0.25) = 14.25 crashes expected$$

6. Prevented

   $$19 - 14.25 = 4.75 crashes over 3 years$$

Why the other options are there

• 4.22 (crashes per mile, no exposure)
• 14.3 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 1.5 mile segment with an ADT of 7,562 vehicles per day experienced 47 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

   $$ADT \times 365 \times N \times L = 7562 \times 365 \times 5 \times 1.5 = 2.070e+7 veh-mi$$

3. Substituting

   $$R = (47 \times 10^{8})/2.070e+7 = 227.0 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$47(1 - 0.30) = 32.90 crashes expected$$

6. Prevented

   $$47 - 32.90 = 14.10 crashes over 5 years$$

Why the other options are there

• 31.33 (crashes per mile, no exposure)
• 70.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 3.5 mile segment with an ADT of 44,031 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

   $$ADT \times 365 \times N \times L = 44031 \times 365 \times 5 \times 3.5 = 2.812e+8 veh-mi$$

3. Substituting

   $$R = (58 \times 10^{8})/2.812e+8 = 20.62 crashes per 100 MVM$$

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

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

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections

A 1.5 mile segment with an ADT of 32,569 vehicles per day experienced 51 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

   $$ADT \times 365 \times N \times L = 32569 \times 365 \times 5 \times 1.5 = 8.916e+7 veh-mi$$

3. Substituting

   $$R = (51 \times 10^{8})/8.916e+7 = 57.20 crashes per 100 MVM$$

4. Formula

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

5. Substituting

   $$51(1 - 0.40) = 30.60 crashes expected$$

6. Prevented

   $$51 - 30.60 = 20.40 crashes over 5 years$$

Why the other options are there

• 34.00 (crashes per mile, no exposure)
• 102.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates at Intersections