Question

a driver in a car, originally moving at 11.8 m/s, applies the brakes until the car comes to a stop. the car moves a distance of 36.3 m while braking. how much time did it take for the car to stop? assume constant acceleration during braking.

Explanation:

Step1: Recall the kinematic formula for average velocity

The average velocity \( v_{avg} \) for an object with constant acceleration is given by the average of the initial velocity \( v_0 \) and final velocity \( v \). Since the car comes to a stop, \( v = 0 \, \text{m/s} \), \( v_0 = 11.8 \, \text{m/s} \). So \( v_{avg}=\frac{v_0 + v}{2}=\frac{11.8 + 0}{2}= 5.9 \, \text{m/s} \)

Step2: Use the formula for distance in terms of average velocity and time

The formula relating distance \( d \), average velocity \( v_{avg} \), and time \( t \) is \( d=v_{avg}\times t \). We know \( d = 36.3 \, \text{m} \) and \( v_{avg}=5.9 \, \text{m/s} \). Rearranging for \( t \), we get \( t=\frac{d}{v_{avg}} \)

Step3: Substitute the values to find time

Substitute \( d = 36.3 \, \text{m} \) and \( v_{avg}=5.9 \, \text{m/s} \) into the formula: \( t=\frac{36.3}{5.9}\approx6.15 \, \text{s} \)

Answer:

\( 6.15 \) (or more precise value depending on calculation, if we do \( 36.3\div5.9 = 6.1525\cdots\approx6.15 \) or using another kinematic formula \( v^2=v_0^2 + 2ad \) to find acceleration first: \( 0=(11.8)^2+2a\times36.3\Rightarrow a=\frac{- 11.8^2}{2\times36.3}=\frac{-139.24}{72.6}\approx - 1.918 \, \text{m/s}^2 \), then use \( v = v_0+at\Rightarrow0 = 11.8-1.918t\Rightarrow t=\frac{11.8}{1.918}\approx6.15 \, \text{s} \))