# Vectors - Displacement to Accelleration

1. Nov 17, 2011

### cryodynamics

1. The problem statement, all variables and given/known data
An object moves with its position obeying x=5t^2-6t^2 and y=3t^-1+4t^4. What is the magnitude of its acceleration at t=1s;
a)57 m/s^2
b)34 m/s^2
c)26 m/s^2
d)44 m/s^2
e)0 m/s^2

2. Relevant equations
x=5t^3-6t^2
y=3t^-1+4t^4

3. The attempt at a solution
Second derivative of the displacement x is;
ax=30t-12 m/s^2
Second derivative of the displacement y is;
ay=8t m/s^2

t=1 -> ax
ax = 18

t=1 -> ay
ay = 8

Resultant acceleration = sqrt(18^2+8^2) = 19.69

Which is wrong according to the answer I found online which was a) 57 m/s^2. Anyone show me where I've messed up or is the answer I found wrong?

Last edited: Nov 17, 2011
2. Nov 17, 2011

### PeterO

The x expressions don't match up, but I guess you actually meant x=5t^3-6t^2

I can't imagine what you thought the y expression was ??

3. Nov 17, 2011

### cryodynamics

Ahh yes sorry about that, I'll fix that up. What do you mean by 'I can't imagine what you thought the y expression was?' That equation for y is the exact one given?

4. Nov 17, 2011

### grzz

Can you check that

x=5t^2-6t^2 and y=3t^-1+4t^4

are correct.

5. Nov 17, 2011

### cryodynamics

x=5t3-6t2
y=3t-1+4t4
Are the the correct equations given in the question.

6. Nov 17, 2011

### PeterO

OK .. if that is the expression for y, then 8t is not the second derivative.

7. Nov 17, 2011

### cryodynamics

Ahh I got the answer this time, my problem was I thought I could simplify y=3t-1+4t4 to 4t3/3 I now realize what a rookie error that was lol. Thanks for your help everyone.