Question

ms. sardo is driving along with a negative velocity and a positive acceleration in the image to the right. the distance between points 0 and 1 is the same as the distance between points 1 and 2. compare the amount of time it takes ms. sardo to get from 0 to 1 and the time it takes her to get from 1 to 2.
a) it takes longer to get from 0 to 1
b) it takes longer to get from 1 to 2
c) it takes the same amount of time for each interval
d) more information is needed to determine

Explanation:

Step1: Analyze the motion

Since velocity \(v\) and acceleration \(a\) have opposite signs (\(v<0,a > 0\)), the car is decelerating. The average velocity \(v_{avg}=\frac{v_{initial}+v_{final}}{2}\)

Let the distance between \(0 - 1\) and \(1 - 2\) be \(d\). For the motion from \(0\) to \(1\): Let the initial velocity be \(v_0\) (negative) and the final velocity be \(v_1\) (more negative than \(v_0\) because \(a>0\) but \(v<0\)). So, \(v_{avg1}=\frac{v_0 + v_1}{2}\)

For the motion from \(1\) to \(2\): The initial velocity is \(v_1\) (more negative) and the final velocity is \(v_2\) (even more negative). So, \(v_{avg2}=\frac{v_1 + v_2}{2}\), and \(|v_{avg2}|<|v_{avg1}|\)

Step2: Use the formula \(t=\frac{d}{v_{avg}}\)

Since \(d\) is the same for both intervals (\(d_1 = d_2=d\)), and \(|v_{avg2}|<|v_{avg1}|\) (because the car is decelerating and velocity is negative). Using the formula \(t=\frac{d}{v_{avg}}\), when \(d\) is constant, \(t\) is inversely proportional to \(|v_{avg}|\).

Since \(|v_{avg2}|<|v_{avg1}|\), \(t_2=\frac{d}{|v_{avg2}|}\) and \(t_1=\frac{d}{|v_{avg1}|}\), so \(t_2>t_1\)

Answer:

B. It takes longer to get from 1 to 2