Question

kala is driving a racecar. the table below gives the distance d(t) (in meters) she has driven at a few times t (in seconds) after she starts.

time t (seconds)	distance d(t) (meters)
3	78.3
4	147.6
6	185.4
9	287.1

(a) find the average rate of change for the distance driven from 0 seconds to 4 seconds.

meters per second

(b) find the average rate of change for the distance driven from 6 seconds to 9 seconds.

meters per second

Explanation:

Step1: Recall average rate of change formula

The average rate of change of a function \( D(t) \) from \( t = a \) to \( t = b \) is given by \( \frac{D(b) - D(a)}{b - a} \).

Step2: Solve part (a)

For \( a = 0 \) and \( b = 4 \):

• \( D(0) = 0 \) meters, \( D(4) = 147.6 \) meters.
• Substitute into the formula: \( \frac{D(4) - D(0)}{4 - 0} = \frac{147.6 - 0}{4 - 0} \)
• Calculate: \( \frac{147.6}{4} = 36.9 \)

Step3: Solve part (b)

For \( a = 6 \) and \( b = 9 \):

• \( D(6) = 185.4 \) meters, \( D(9) = 287.1 \) meters.
• Substitute into the formula: \( \frac{D(9) - D(6)}{9 - 6} = \frac{287.1 - 185.4}{9 - 6} \)
• Calculate numerator: \( 287.1 - 185.4 = 101.7 \)
• Calculate denominator: \( 9 - 6 = 3 \)
• Divide: \( \frac{101.7}{3} = 33.9 \)

Answer:

(a) \( 36.9 \) meters per second
(b) \( 33.9 \) meters per second