Question

a distracted driver going 33km/h suddenly sees a telephone pole 30.6 m ahead. what braking acceleration is necessary to stop in time (in m/s²)?

Explanation:

Step1: Convert the initial velocity unit

The initial velocity \(v_0 = 33\space km/h\). Convert it to \(m/s\): \(v_0=\frac{33\times1000}{3600}\space m/s=\frac{330}{36}\space m/s=\frac{55}{6}\space m/s\approx9.17\space m/s\). The final velocity \(v = 0\space m/s\), and the displacement \(x = 30.6\space m\).

Step2: Use the kinematic equation \(v^{2}=v_{0}^{2}+2ax\)

Rearrange the equation \(v^{2}=v_{0}^{2}+2ax\) to solve for \(a\). We get \(a=\frac{v^{2}-v_{0}^{2}}{2x}\).
Substitute \(v = 0\), \(v_0=\frac{55}{6}\space m/s\), and \(x = 30.6\space m\) into the formula:

$$
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Answer:

The braking acceleration is approximately \(-1.37\space m/s^{2}\)