A car moves along a straight road in such a way that its velocity (in feet per second) at any time t (in seconds) is given by

v(t) = 3t sqrt(64−t^2)
(0 ≤ t ≤ 8).
Find the distance traveled by the car in the 8 sec from t = 0 to t = 8.

distance is

∫[0,8] 3t√(640t^2) dt
= 512

To find the distance traveled by the car in the 8 seconds from t = 0 to t = 8, we need to integrate the absolute value of the velocity function over the given time interval.

The velocity function, v(t), is given as 3t sqrt(64−t^2) for 0 ≤ t ≤ 8.

To find the distance, we need to remove the negative values from the velocity function and integrate the absolute value of the velocity over the interval.

1. First, let's rewrite the velocity function as:
v(t) = 3t √(64−t^2)

2. Since we only need the positive values of velocity, we take the absolute value of the function:
|v(t)| = |3t √(64−t^2)|

3. The distance traveled is obtained by integrating the absolute value of the velocity function from t = 0 to t = 8:
distance = ∫[0 to 8] |3t √(64−t^2)| dt

4. To integrate the absolute value, we need to consider the positive and negative parts separately:
distance = ∫[0 to 8] (3t √(64−t^2)) dt - ∫[0 to 8] (-3t √(64−t^2)) dt

5. Simplifying the integrals:
distance = ∫[0 to 8] (3t √(64−t^2)) dt + ∫[0 to 8] (3t √(64−t^2)) dt

6. Combining the integrals:
distance = 2 ∫[0 to 8] (3t √(64−t^2)) dt

7. Now, to evaluate the integral, we can use a substitution. Let's substitute u = 64−t^2.

8. Differentiating both sides of the equation, we get du = -2t dt.
Rearranging, dt = -(1/2t) du.

9. Substituting these values back into the integral:
distance = 2 ∫[0 to 8] (3t √(u)) (-1/2t) du
distance = - ∫[0 to 8] 3√(u) du

10. We can simplify the integral now:
distance = -3 ∫[0 to 8] u^(1/2) du

11. Integrating u^(1/2), we get:
distance = -3 [ (2/3)u^(3/2) ] [0 to 8]
distance = -3 [ (2/3)(64-0)^(3/2) ]
distance = -3 (2/3) (8^3/2)
distance = -3 (2/3) (8√8)
distance = -16√8
distance = -16 * 2√2
distance = -32√2
(Note: We disregard the negative sign as the distance traveled cannot be negative.)

Therefore, the distance traveled by the car in the 8 seconds is 32√2 feet.