Since acceleration is the derivative of velocity, you have to take the slope between 15 and 29.
f(b)-f(a)/b-a
(87-45)/(29-15)=42/14= 3 m/s2
time Velocity
seconds (m/s)
x f (x)
15 45
22 66
29 87
36 108
The average rate of change of the function between x = 15 to x = 29 is __________m/s2 and represents the car's acceleration
f(b)-f(a)/b-a
(87-45)/(29-15)=42/14= 3 m/s2
f(b)-f(a)/b-a
(87-45)/(29-15)=42/14= 3 m/s2
That means that 3m/s2 is 3 meters per second
Change in velocity = final velocity - initial velocity = f(29) - f(15) = 87 - 45 = 42 m/s
Change in time = final time - initial time = 29 - 15 = 14 seconds
Average rate of change = change in velocity / change in time = 42 m/s / 14 s = 3 m/sĀ²
Therefore, the average rate of change of the function between x = 15 to x = 29 is 3 m/sĀ², which represents the car's acceleration.
Step 1: Find the change in velocity.
The change in velocity is obtained by subtracting the initial velocity from the final velocity. In this case, the initial velocity is the velocity at x = 15, which is 45 m/s, and the final velocity is the velocity at x = 29, which is 87 m/s.
Change in velocity = Final velocity - Initial velocity
Change in velocity = 87 m/s - 45 m/s
Change in velocity = 42 m/s
Step 2: Find the change in time.
The change in time is obtained by subtracting the initial time from the final time. In this case, the initial time is 15 seconds, and the final time is 29 seconds.
Change in time = Final time - Initial time
Change in time = 29 s - 15 s
Change in time = 14 s
Step 3: Calculate the average rate of change (acceleration).
Average rate of change (acceleration) = Change in velocity / Change in time
Average rate of change (acceleration) = 42 m/s / 14 s
Average rate of change (acceleration) = 3 m/s^2
Therefore, the average rate of change of the function between x = 15 to x = 29 is 3 m/s^2, which represents the car's acceleration.