A particle moves along the c-axis so that at time t its position is given by x(t)=t^2-6^t+9t+11

a)What is the velocity of the particle at t=0
b)During what time intervals is the particle moving to the left?
c)What is the total distance traveled by the particle t=0 to t=2

I can't believe that is the correct equation of motion. 6^t will grow negative quite quickly. Since there's a typo in c-axis, I suspect there's another in the equation.

opps it is x(t)=t^3-6t^2+9t+11

Sorry!!

That's better!

x(t) is the position
x'(t) is the velocity

x'(t) = 3t^2 - 12t + 9
x'(0) = 9

particle moving to the left means velocity is negative.

3(t^2 - 4t + 3) = 3(t-1)(t-3) < 0
when 1 < t < 3

distance traveled is x(2) - x(0)

thanks!!! that was VERY helpful!!

To find the velocity of the particle at a specific time t, we need to find the derivative of the position function x(t) with respect to time. The velocity function, v(t), is the derivative of x(t).

a) Velocity at t=0:
To find the velocity at t=0, we differentiate x(t) with respect to t:
x(t) = t^2 - 6t + 9t + 11
v(t) = d/dt (t^2 - 6t + 9t + 11)
= 2t - 6 + 9
= 2t + 3

Now substitute t=0 into the velocity function:
v(0) = 2(0) + 3
= 0 + 3
= 3

Therefore, the velocity of the particle at t=0 is 3.

b) Time intervals when the particle is moving to the left:
To determine when the particle is moving to the left, we need to find when the velocity function is negative. The particle moves to the left when its velocity is negative.

From part a), we found that v(t) = 2t + 3. Setting v(t) < 0:
2t + 3 < 0

Solving for t:
2t < -3
t < -3/2

Hence, the particle is moving to the left when t is less than -3/2.

c) Total distance traveled from t=0 to t=2:
The total distance traveled by the particle can be calculated by finding the definite integral of the absolute value of the velocity function over the interval [0, 2].

First, find the absolute value of the velocity function:
|v(t)| = |2t + 3|

Now integrate the absolute value of the velocity function over the interval [0, 2]:
Integral from 0 to 2 of |v(t)| dt

To compute this integral, we need to split the integral into two separate intervals:

1) When v(t) is positive (moving to the right):
Integral from 0 to 2 of (2t + 3) dt

2) When v(t) is negative (moving to the left):
Integral from 0 to -3/2 of -(2t + 3) dt

Evaluate both integrals to find the respective areas under the curve.

Add the absolute values of the results from the two integrals to get the total distance traveled by the particle from t=0 to t=2.