# Homework Help: [DYNAMICS] Acceleration as a function of position.

1. May 30, 2013

### Brevity

1. The problem statement, all variables and given/known data

Initial velocity = 17 m/s
Radius of curvature = 13 m
Acceleration as a function of position = -0.25s m/s^2
Time elapsed = 2 s

Solve for the magnitude of the acceleration.

2. Relevant equations

a = v (dv/ds)

a^2 = a(t)^2 + a(n)^2, wherein a(t) is tangential acceleration and a(n) is normalized acceleration.

3. The attempt at a solution

a*ds = v*dv

int(-0.25s*ds) = int(v*dv)

v^2 = -0.25s^2

From this point I have hit significant difficulty, as you cannot take the square root of a negative number. Presumably there is another method to solve this equation that I have neglected, but it hasn't occurred to me, and my professor is notoriously poor at returning emails. Any thoughts?

2. May 30, 2013

### rcgldr

Shouldn't that be

1/2 v^2 = c - 0.25s^2

where c is a constant of integration. Assuming c is greater than .25s^2, then 1/2 v^2 is positive.

Also, you might consider using another name for position or distance, to avoid confusion since s is often use to represent time in seconds.

1/2 v^2 = c - 0.25 d^2

3. May 30, 2013

### Brevity

Understood. It would then be 1/2 v^2 = C - 0.125 d^2, would it not? Let me try it out.

4. May 30, 2013

### Brevity

Unless I have misinterpreted something horribly, this was my process:

s denotes position.
t denotes time.

-0.25s ds = v dv
C -0.125s^2 = 1/2 v^2
When s = 0, v = 17.
C = 1/2 (17)^2
C = 144.5
144.5 - 0.125s^2 = 1/2v^2
289 - 0.25s^2 = v^2
v = sqrt(289 - 0.25s^2)
ds / dt = sqrt(289 - 0.25s^2)
(289 - 0.25s^2)^(-1/2) ds = dt
2*sin^-1(0.029412s) = t
t = 2
s = 34sin(1)
s = 28.61

v = sqrt(289 - 0.25s^2)
v = sqrt(84.367)
v = 9.185

a(n) = v^2 / ρ
v = 9.185
ρ = 13
a(n) = 6.48956

a(t) = -0.25s
s = 28.61
a(t) = -7.1525

a = sqrt(a(n)^2+a(t)^2)
a = sqrt(42.114+51.1583)
a = 9.66