How Do You Calculate the Magnitude of Acceleration for a Particle in Motion?

[schrock]

A location of a particle is given in m by x, y and z coordinates as function of time in s as:
x= -11+9t+11t^2
y= -23-21t
z= -93+25t+11t^2

What is the magnitude of the objects acceleration at t= 3.00s?

Would I add the second derivatives for magnitude?

 

[tiny-tim]

Hi schrock!

A location of a particle is given in m by x, y and z coordinates as function of time in s as:
x= -11+9t+11t^2
y= -23-21t
z= -93+25t+11t^2

What is the magnitude of the objects acceleration at t= 3.00s?

Would I add the second derivatives for magnitude?

You mean |a| = x'' + y'' + z'' ?

Nooo …

Acceleration is a vector (just like velocity ).

So its magnitude is calculated from its components the same way as for any vector.

 

[baba89]

Yes, to find the magnitude of acceleration at a specific time, you would need to calculate the second derivatives of the position equations with respect to time and then use the Pythagorean theorem to find the magnitude of the resulting acceleration vector. In this case, the magnitude of acceleration at t=3.00s would be calculated as follows:

x''(3.00) = 18
y''(3.00) = -21
z''(3.00) = 36

Magnitude of acceleration = √(x''(3.00)^2 + y''(3.00)^2 + z''(3.00)^2)
= √(18^2 + (-21)^2 + 36^2)
= √(324 + 441 + 1296)
= √2061
= 45.41 m/s^2

Therefore, the magnitude of acceleration at t=3.00s is 45.41 m/s^2.