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)?

-1.37 m/s²
-0.0562 m/s²
-17.8 m/s²
-0.157 m/s²

Explanation:

Step1: Convert the initial velocity to m/s

The initial velocity \(v_0 = 33\ km/h\). To convert it to \(m/s\), use the conversion factor \(1\ km = 1000\ m\) and \(1\ h=3600\ s\).
\(v_0=\frac{33\times1000}{3600}\ m/s=\frac{33000}{3600}\ m/s = 9.1667\ m/s\)
The final velocity \(v = 0\ m/s\) (the car stops), and the displacement \(x - x_0=30.6\ m\)

Step2: Use the kinematic equation \(v^{2}=v_{0}^{2}+2a(x - x_0)\)

Rearrange the equation \(v^{2}=v_{0}^{2}+2a(x - x_0)\) for \(a\).
\(a=\frac{v^{2}-v_{0}^{2}}{2(x - x_0)}\)
Substitute \(v = 0\), \(v_0 = 9.1667\ m/s\), and \(x - x_0 = 30.6\ m\) into the formula.
\(a=\frac{0-(9.1667)^{2}}{2\times30.6}\)
First, calculate \((9.1667)^{2}=84.0278\)
Then \(2\times30.6 = 61.2\)
\(a=\frac{- 84.0278}{61.2}=-1.37\ m/s^{2}\)

Answer:

\(-1.37\ m/s^{2}\) (the first option)