Acceleration with position given

[silentsaber]

Homework Statement

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?

Homework Equations

v=x/t a=v/t

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?

 

[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.

 

[nlightner]

Hello,

Your attempt at a solution is on the right track, but there are a few errors in your calculations. Firstly, the hypotenuse of the triangle formed by the x and y coordinates is not always 4. It varies depending on the value of t. To find the magnitude of acceleration, we need to use the equation a = v/t, where v is the velocity and t is the time.

To find the velocity, we can take the derivative of the position function r(t). This will give us the velocity function v(t). So, v(t) = -12 sin(3t) i + 12 cos(3t) j. Now, we can plug in the value of t (30 seconds) to find the velocity at that point. However, note that the velocity is a vector quantity, so we need to find the magnitude of this vector using the Pythagorean theorem.

Once we have the magnitude of velocity, we can divide it by the time (30 seconds) to find the acceleration at that point. This will give us the correct answer.

I hope this helps! Let me know if you have any further questions.