Question

during the first 8.00 seconds after a car started moving, its speed increased to 11.0 meters per second. from 8.00 seconds to 14 seconds after the car started moving, its speed increased from 11.0 meters per second to 23.0 meters per second. to the nearest hundredth, what is the positive difference between the average rate of change of speed of the car, in meters per second, during the first 8.00 seconds after the car started moving and the average rate of change of the speed of the car, in meters per second, from 8.00 seconds to 14.0 seconds after the car started moving?

Explanation:

Step1: Calculate average rate of change for first 8 seconds

The formula for average rate of change (which is acceleration here) is $\frac{\Delta v}{\Delta t}$. For the first 8 seconds, initial speed $v_1 = 8.00$ m/s, final speed $v_2 = 11.0$ m/s, time $\Delta t_1 = 8.00$ s. So, $a_1=\frac{11.0 - 8.00}{8.00}=\frac{3.00}{8.00}= 0.375$ m/s².

Step2: Calculate average rate of change for 8 - 14 seconds

For the time from 8.00 s to 14.0 s, initial speed $v_3 = 11.0$ m/s, final speed $v_4 = 23.0$ m/s, time $\Delta t_2=14.0 - 8.00 = 6.00$ s. So, $a_2=\frac{23.0 - 11.0}{6.00}=\frac{12.0}{6.00}=2.00$ m/s².

Step3: Find the positive difference

Subtract the two average rates of change: $|a_2 - a_1|=|2.00 - 0.375| = 1.625$ m/s². Rounding to the nearest hundredth gives $1.63$ m/s².

Answer:

1.63