A car drives down a road in such a way that its velocity (in m/s) at time t (seconds) is v(t)=3(t^(1/2))+4. Find the car's average velocity (in m/s) between t=4 and t=6.

find out how far it went between t = 4 and t = 6, then divide by 2

int [ 3 t^.5]dt + int [4]dt t=4 to t=6

3(2/3)t^3/2 + 4t

2 t^3/2 + 4 t

at t = 6 , 2 sqrt(6^3) + 24=
6sqrt6+24
=38.7

at t = 4, 2 sqrt(4^3) + 16=
2*8 + 16 = 32

38.7 - 32

then divide by two seconds to find average m/s

(38.7-32)/2=3.35

It's not the correct answer.

To find the car's average velocity between t=4 and t=6, we need to calculate the total displacement and divide it by the time interval.

The total displacement can be found by calculating the definite integral of the velocity function v(t) between t=4 and t=6. The definite integral of v(t) gives us the displacement function s(t).

Let's calculate the displacement function first:

Step 1: Integrate the velocity function v(t) with respect to t.
∫(v(t)) dt = ∫(3(t^(1/2)) + 4) dt = ∫3(t^(1/2)) dt + ∫4 dt

Step 2: Evaluate the indefinite integrals:
= 2(t^(3/2)) + 4t + C1 + 4t + C2
= 2(t^(3/2)) + 8t + C

Step 3: Evaluate the definite integral between t=4 and t=6:
s(t=6) - s(t=4) = 2(6^(3/2)) + 8(6) + C - (2(4^(3/2)) + 8(4) + C)
= 2(36^(1/2)) + 48 - (2(16^(1/2)) + 32)
= 2(6) + 48 - (2(4) + 32)
= 12 + 48 - (8 + 32)
= 60 - 40
= 20

The total displacement is 20 meters.

Now, to find the average velocity, we divide the displacement by the time interval:
Average velocity = Total displacement / Time interval
= 20 / (6 - 4)
= 20 / 2
= 10 m/s

Therefore, the car's average velocity between t=4 and t=6 is 10 m/s.