# Thread: Rectilinear motion using Antiderivatives

1. ## Rectilinear motion using Antiderivatives

If the brakes on a car can give the car a constant negative acceleration of 8m/sec^2, what is the greatest speed it may be going if it is necessary to be able to stop the car within 25m after the brake is applied?

20m/sec = 72km/hr

Helppp *cough cough* ....

2. Originally Posted by ^_^Engineer_Adam^_^
If the brakes on a car can give the car a constant negative acceleration of 8m/sec^2, what is the greatest speed it may be going if it is necessary to be able to stop the car within 25m after the brake is applied?

20m/sec = 72km/hr

Helppp *cough cough* ....
We want the greatest speed, so let's assume a stopping distance of 25 m. Set the origin of the problem to be where the car starts to apply the brakes and take a positive x direction to be the direction the car is initially moving in. So:
..........t = ?
x0 = 0 m.......x = 25 m
v0 = ?..........v = 0 m/2 (since it has stopped)
..........a = -8 m/s^2

We want to know v0, so we need to know an equation for v0 that doesn't contain t. A good choice would be:
v^2 = v0^2 + 2a(x - x0)

0 = v0^2 +2(-8)(25 - 0) <-- Leaving the units out for convenience.

0 = v0^2 - 400

v0^2 = 400

v0 = (+/-) 20 m/s

Now, which sign do we choose? Since all we want is the speed, which is the magnitude of the velocity, we don't care. (For the record, the initial velocity is positive, since that's how we defined the positive direction in the first place.) So the greatest speed is 20 m/s.

-Dan

Another approach . . .

If the brakes on a car can give the car a constant acceleration of -8 m/sec²,
what is the greatest speed it may be going if it is necessary to be able
to stop the car within 25m after the brake is applied?

Acceleration is the derivative of the velocity function.

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

. . Integrate: .v(t) .= .-8t + v
0

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

Velocity is the derivative of the position function.

We have: .v(t) .= .x'(t) .= .-8t + v
0

Integrate: .x(t) .= .-4t² + v
0t + x0

Assume that x
0 = 0.
. . We have: .x(t) .= .-4t² + v
0t

If x = 25, we have: .-4t² + v
0t .= .25 . . 4t² - v0t + 25 .= .0
. . . . . . . . . . . . . . . . . . . . . . .________
. . . . . . . . . . . . . . . . . . v
0 ± √v0² - 400
Quadratic Formula: . t .= .-------------------
. . . . . . . . . . . . . . . . . . . . . . . 8

. . . . . . . . . . . . . . . . ________
. . . . . . . . . . . .v
0 ± √v0² - 400 . . . . .v0
Equate to [1]: . -------------------- . = . ---
. . . . . . . . . . . . . . . . .8 . . . . . . . . . .8
. . . . . . .________
Then: . √v
o² - 400 .= .0 . . v0² - 400 .= .0 . . v0² .= .400

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

4. Originally Posted by Soroban

Another approach . . .

Acceleration is the derivative of the velocity function.

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

. . Integrate: .v(t) .= .-8t + v
0

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

Velocity is the derivative of the position function.

We have: .v(t) .= .x'(t) .= .-8t + v
0

Integrate: .x(t) .= .-4t² + v
0t + x0

Assume that x
0 = 0.
. . We have: .x(t) .= .-4t² + v
0t

If x = 25, we have: .-4t² + v
0t .= .25 . . 4t² - v0t + 25 .= .0
. . . . . . . . . . . . . . . . . . . . . . .________
. . . . . . . . . . . . . . . . . . v
0 ± √v0² - 400
Quadratic Formula: . t .= .-------------------
. . . . . . . . . . . . . . . . . . . . . . . 8

. . . . . . . . . . . . . . . . ________
. . . . . . . . . . . .v
0 ± √v0² - 400 . . . . .v0
Equate to [1]: . -------------------- . = . ---
. . . . . . . . . . . . . . . . .8 . . . . . . . . . .8
. . . . . . .________
Then: . √v
o² - 400 .= .0 . . v0² - 400 .= .0 . . v0² .= .400

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

Hmmmm...that's the hard way, but given the title of his post it's probably what he was after...

-Dan