Question

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Explanation:

First Problem (Deceleration Lane Length)

Step1: Identify the formula for distance with constant acceleration (average velocity formula)

The formula for the distance \(\Delta d\) traveled when velocity changes from initial velocity \(v_i\) to final velocity \(v_f\) in time \(\Delta t\) is \(\Delta d=\frac{v_i + v_f}{2}\times\Delta t\). Here, \(v_i = 35\space m/s\), \(v_f=18\space m/s\), and \(\Delta t = 8.0\space s\).

Step2: Substitute the values into the formula

Substitute \(v_i = 35\), \(v_f = 18\), and \(\Delta t=8.0\) into the formula:

$$ \Delta d=\frac{35 + 18}{2}\times8.0=\frac{53}{2}\times8.0 = 53\times4.0=212\space m $$

Step3: Express in scientific notation

\(212\space m\) can be written as \(2.12\times 10^{2}\space m\). So, \(a = 2\), \(b = 1\) (or \(2\) if we consider significant figures, but based on the calculation \(212 = 2.12\times 10^{2}\)), and \(c = 2\).

Step1: Convert initial velocity to m/s

The initial velocity \(v_i=105\space km/h\). To convert to \(m/s\), use the conversion factor \(1\space km = 1000\space m\) and \(1\space h=3600\space s\). So, \(v_i=\frac{105\times1000}{3600}\approx29.2\space m/s\), \(v_f = 0\space m/s\) (stops), \(\Delta t = 3.8\space s\).

Step2: Use the average velocity formula

Using \(\Delta d=\frac{v_i + v_f}{2}\times\Delta t\), substitute \(v_i\approx29.2\), \(v_f = 0\), \(\Delta t = 3.8\):

$$ \Delta d=\frac{29.2+0}{2}\times3.8 = 14.6\times3.8 = 55.48\approx55\space m $$

(If we use more precise calculation \(\frac{105}{3.6}=29.1667\space m/s\), then \(\Delta d=\frac{29.1667 + 0}{2}\times3.8=14.58335\times3.8 = 55.41673\approx55\space m\) or \(5.5\times 10^{1}\space m\))

Step1: Use the distance formula

We know \(\Delta d=\frac{v_i + v_f}{2}\times\Delta t\), \(\Delta d = 75.0\space m\), \(\Delta t=10.0\space s\), and \(v_f = 0\space m/s\) (comes to stop). Substitute these values:

$$ 75.0=\frac{v_i+0}{2}\times10.0 $$

Step2: Solve for \(v_i\)

First, simplify the equation:

$$ 75.0=\frac{v_i\times10.0}{2}=v_i\times5.0 $$

Then, solve for \(v_i\):

$$ v_i=\frac{75.0}{5.0}=15\space m/s $$

Step3: Convert to km/h

To convert \(m/s\) to \(km/h\), use the conversion factor \(1\space m/s=3.6\space km/h\). So, \(v_i = 15\times3.6 = 54\space km/h\).

Answer:

\(a = 2\), \(b = 1\) (or \(2\)), \(c = 2\) (The most accurate from calculation is \(a = 2\), \(b=1\) (if we take \(2.12\) then \(b = 1\)), \(c = 2\))

Example 2 (Braking Distance)