#### To determine

**To estimate:**

The increase in the velocity of the car during the six-second time interval.

#### Answer

39.8 ft/s

#### Explanation

**1) Concept:**

Use the Net Change theorem and the Midpoint Rule.

**2) Theorem and Rule:**

Net Change theorem: The integral of a rate of change is the net change.

∫abF'(x)dx=Fb-F(a)

The Midpoint Rule:

∫abfxdx≈∑i=1nfxi- ∆x=∆x fx1-+…+fxn-

where ∆x=b-an and xi-=12xi-1+xi midpoint of xi-1, xi.

**2) Calculation:**

Acceleration is the rate of change of velocity v with respect to time t.

at=v'(t)

Here t∈0,6

By Net Change theorem,

∫06at dt≈∫06v'tdt=v6-v(0)

is the change in velocity of the car during the given time interval.

Approximate the value of the integral using the Midpoint Rule with three subintervals: that is with n=3, a=0 and b=6 So width of the interval is

∆t=b-an

Substitute values.

∆t=6-03

∆t=2

From the graph, the three subintervals are 0,2, 2,4, 4,6 and the midpoints are

t1-=12t0+t2=120+2=122=1

t2-=12t2+t4=122+4=126=3

t3-=12t4+t6=124+6=1210=5

So, the value of integral is

∫06at dt≈M6=∑i=13ati- ∆t

=∆tat1-+at2-+at3-

=∆ta1+a3+a5

From the given graph,

a1=0.6

a3=10

a5=9.3

=∑i=13axi- ∆x≈0.6+10+9.32=39.8 ft/s

**Conclusion:**

Therefore, the increase in the velocity of the car during the six-second time interval is 39.8 ft/s.