Question

the position function of a particle in rectilinear motion is given by s(t)=t^3-9t^2+24t+1 for t greater than or equal to 0 with t measured in seconds and s(t) measured in feet. find position and acceleration of the particle at the instant when the particle reverses direction

Answer 1

The position of the particle at this instant is 13 feet.

The acceleration at this instant is -6 feet per second squared, ( that is deceleration).

Step-by-step explanation:

To find the position and acceleration of the particle at the instant when the particle reverses direction, we need to find the time when the particle's velocity becomes zero.

We can do this by finding the value of t for which the derivative of the position function, s(t), is equal to zero.

Taking the derivative of s(t), we get v(t) = 3t^2 - 18t + 24.

Setting v(t) equal to zero and solving for t, we find that the particle's velocity becomes zero at t = 2 seconds.

To find the position of the particle at this instant, we substitute t = 2 into the position function:

s(2) = (2)^3 - 9(2)^2 + 24(2) + 1

= 13 feet.

To find the acceleration of the particle at the instant when it reverses direction, we can take the derivative of the velocity function, a(t).

The derivative of v(t) is a(t) = 6t - 18.

Substituting t = 2 into a(t), we find that the particle's acceleration at the instant it reverses direction is

a = 6(2) - 18 = -6 feet per second squared.