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Answer

b

15.18 m\text{15.18 m}

44.625 m\text{44.625 m}

74.26 km/hr\text{74.26 km/hr}

Step-by-step solution

15 steps

Part a

To determine if the driver will be arrested, we need to check their speed. Police arrest drivers traveling above 80 km/hr80 \text{ km/hr}.

The speed for the first part of the journey was 60 km/hr60 \text{ km/hr}.
The speed for the second part of the journey was 80 km/hr80 \text{ km/hr}.

Since neither speed is above 80 km/hr80 \text{ km/hr} (60 km/hr≯80 km/hr60 \text{ km/hr} \ngtr 80 \text{ km/hr} and 80 km/hr≯80 km/hr80 \text{ km/hr} \ngtr 80 \text{ km/hr}), the driver will not be arrested.

The driver will not be arrested.

Part b

15.18 m\text{15.18 m}

We use the stopping distance formula d=s2200+s10d = \frac{s^2}{200} + \frac{s}{10}.

To find the stopping distance for s=46 km/hrs = 46 \text{ km/hr}:

  1. Substitute s=46s = 46 into the formula.

    d=462200+4610d = \frac{46^2}{200} + \frac{46}{10}

  2. Calculate 46246^2.

    d=2116200+4610d = \frac{2116}{200} + \frac{46}{10}

  3. Perform the divisions.

    d=10.58+4.6d = 10.58 + 4.6

  4. Add the values.

    d=15.18 md = 15.18 \text{ m}

    The stopping distance for 46 km/hr46 \text{ km/hr} is 15.18 m\boxed{\text{15.18 m}}.

    To find the stopping distance for s=85 km/hrs = 85 \text{ km/hr}:

  5. Substitute s=85s = 85 into the formula.

    d=852200+8510d = \frac{85^2}{200} + \frac{85}{10}

  6. Calculate 85285^2.

    d=7225200+8510d = \frac{7225}{200} + \frac{85}{10}

  7. Perform the divisions.

    d=36.125+8.5d = 36.125 + 8.5

  8. Add the values.

    d=44.625 md = 44.625 \text{ m}

    The stopping distance for 85 km/hr85 \text{ km/hr} is 44.625 m\boxed{\text{44.625 m}}.

    To find the speed when the stopping distance is 35 m35 \text{ m}:

  9. Substitute d=35d = 35 into the formula.

    35=s2200+s1035 = \frac{s^2}{200} + \frac{s}{10}

  10. Multiply both sides by 200 to eliminate the denominators.

    35×200=s2+s10×20035 \times 200 = s^2 + \frac{s}{10} \times 200 7000=s2+20s7000 = s^2 + 20s

  11. Rearrange the equation into a standard quadratic form as2+bs+c=0as^2 + bs + c = 0.

    s2+20s−7000=0s^2 + 20s - 7000 = 0

  12. Use the quadratic formula s=−b±b2−4ac2as = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a=1, b=20b=20, c=−7000c=-7000.

    s=−20±202−4(1)(−7000)2(1)s = \frac{-20 \pm \sqrt{20^2 - 4(1)(-7000)}}{2(1)} s=−20±400+280002s = \frac{-20 \pm \sqrt{400 + 28000}}{2} s=−20±284002s = \frac{-20 \pm \sqrt{28400}}{2}

  13. Calculate the square root.

    28400≈168.5229\sqrt{28400} \approx 168.5229

  14. Calculate the two possible values for ss.

    s1=−20+168.52292=148.52292≈74.26145s_1 = \frac{-20 + 168.5229}{2} = \frac{148.5229}{2} \approx 74.26145 s2=−20−168.52292=−188.52292≈−94.26145s_2 = \frac{-20 - 168.5229}{2} = \frac{-188.5229}{2} \approx -94.26145

  15. Choose the positive value for speed, as speed cannot be negative.

    s≈74.26 km/hrs \approx 74.26 \text{ km/hr}

    The speed at which the car is traveling if its stopping distance is 35 m35 \text{ m} is 74.26 km/hr\boxed{\text{74.26 km/hr}}.

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