Magnitude of the Acceleration Particle at the Instant When Velocity is Zero.

[Hckyplayer8]

"A particle moving along the x axis has a position given by x = (24t – 2.0t3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero? Your solution should include equations for both velocity and acceleration as functions of time."

It's the middle portion that is tripping me up.

They give position. The first derivative of position is velocity. The second derivative is acceleration.

Thus

f'(x) = (24-6.0t^2)m

and

f''(x) = (-12t)m

"What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?"

Is this asking me to set velocity to zero, solve for time and then plug that value into the acceleration function?

 

[Dr.Peterson]

Yes.

Do you have an alternative interpretation?

 

[Hckyplayer8]

Yes.

Do you have an alternative interpretation?

Nope. Slowly but surely, I'm building the maths intuition. However, I still need sanity checks. Sometimes all their filler words throw me for a loop and I want to do stupid stuff to the problem.

For example, "What is the magnitude...", so instantly my brain went off in lala-land where I originally thought about throwing the Pythagorean Theorem at this...only to realize it wasn't saying t = 0 but v=0 and then plug that t into acceleration.

Just normal greenhorn stuff.

 

[Steven G]

Remember that a is a vector. That is it has a magnitude and direction. An example of a might be 7?? to the north? What are the units associated with 7? The magnitude is 7.