# Acceleration with position given

1. Mar 22, 2009

### silentsaber

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

The position of an object is given by r(t) = 4 cos(3t) i + 4 sin(3t) j (where distance is
in meters and time is in seconds). What is the magnitude of the object’s acceleration at
any given point in time?

2. Relevant equations

v=x/t a=v/t

3. The attempt at a solution
ok so i thoguht that since the x and y coordinates are 4 cos 3t and 4 sin 3t the hypotenuse would be 4 and then that would be the distance so then i solved for t by 4cos(3t)^2 + 4sin 3t^2=4^2 and i solved t to be 30 since sin of 3(30)=90 is 1 and cos of 90 is 0 then i used the 30 as my time and divided 4/30 to find the velocity and then use the velocity and plugged it in v/t and i get the wrong answer...where did i go wrong?

2. Mar 22, 2009

### CompuChip

v = x / t only gives you the average (which is exact, whenever the acceleration is zero).

If you want the acceleration at any point in time, you need to differentiate.
v(t) = r'(t)
a(t) = v'(t) = r''(t)

where the prime denotes differentiation with respect to time.