Acceleration Problem

[Neoorganboy]

A particle moves along the x-axis so that its position at time t > 0 is given by (x)t and dx/dt = -10t^4 + 9t^2 + 8t. The acceleration of the particle is zero when t= ?

 

[MarkFL]

Hello, and welcome to FMH!

We are given the time rate of change of position (velocity). What relationship exists between velocity and acceleration?

 

[Neoorganboy]

Well, acceleration is the derivative of velocity, correct?
I apologize for such a late reply.

 

[MarkFL]

Well, acceleration is the derivative of velocity, correct?
I apologize for such a late reply.

Yes, acceleration is the time rate of change (derivative with respect to time \(t\)) of velocity. So, can you give the acceleration function?

 

[Neoorganboy]

a(t) = -2t^5 + 3t^3+ 4t^2

 

[MarkFL]

a(t) = -2t^5 + 3t^3+ 4t^2

You've integrated rather than differentiating.

 

[Neoorganboy]

Ah..I see
so,
a(t) =-40t^3 + 18t + 8 ?

 

[Neoorganboy]

If I set that expression to be equal to zero and solve for t, I get 0.831. So then acceleration is zero when t = 0.831. Is this correct?

 

[MarkFL]

Yes, I get the same approximate value.