Originally Posted by

**Soroban** Hello, ^_^Engineer_Adam^_^!

Another approach . . .

Acceleration is the derivative of the velocity function.

We have: .a(t) .= .v'(t) .= .-8

. . Integrate: .v(t) .= .-8t + v0

The car is stopped when v = 0: .-8t + v0 .= .0 . → . t = v0/8 .**[1]**

Velocity is the derivative of the position function.

We have: .v(t) .= .x'(t) .= .-8t + v0

Integrate: .x(t) .= .-4t² + v0t + x0

Assume that x0 = 0.

. . We have: .x(t) .= .-4t² + v0t

If x = 25, we have: .-4t² + v0t .= .25 . → . 4t² - v0t + 25 .= .0

. . . . . . . . . . . . . . . . . . . . . . .________

. . . . . . . . . . . . . . . . . . v0 ± √v0² - 400

Quadratic Formula: . t .= .-------------------

. . . . . . . . . . . . . . . . . . . . . . . 8

. . . . . . . . . . . . . . . . ________

. . . . . . . . . . . .v0 ± √v0² - 400 . . . . .v0

Equate to [1]: . -------------------- . = . ---

. . . . . . . . . . . . . . . . .8 . . . . . . . . . .8

. . . . . . .________

Then: . √vo² - 400 .= .0 . → . v0² - 400 .= .0 . → . v0² .= .400

. . Therefore: . **v****0** .**=** .**20 m/s**