The position of a particle at any time t >= 0 is given by p(t) = -t^3 + t^2 + 5t + 3. How far does the particle travel during the first six seconds?
I calculated the distance traveled between the endpoints of the interval and the points at which the particle changes directions to get the answer 4481/27.
dp/dt = 0 at changes
look for zeros of
-3t^2+2t +5 = 0
t = -1 and t = +5/3
from t = 0 to t = 5/3 velocity is +
p = 3 at 0
p = 6.51 + 3 = 9.51at t = 5/3
so moved +6.51
from t = 5/3 to t = 6 velocity is -
p = 9.51 at 5/3
p = -147 at t = 6
so moved 9.51 -147 = -137.5 from 5/3 to 6
total motion = 6.51 - 137.5 = -131
Shouldn't 9.51 - 147 be 9.51 - (-147)?
Sorry, yes, add the absolute valuess
To find the distance traveled by the particle during the first six seconds, you need to calculate the displacement over the entire interval and then take the absolute value of the result.
The position function p(t) represents the position of the particle at any time t. To find the displacement between two points, you need to subtract the initial position from the final position.
In this case, we want to find the distance traveled during the first six seconds, which means we need to calculate the displacement between t=0 and t=6.
First, we find the initial position by plugging in t=0 into the position function:
p(0) = -(0)^3 + (0)^2 + 5(0) + 3
p(0) = 3
Next, we find the final position by plugging in t=6 into the position function:
p(6) = -(6)^3 + (6)^2 + 5(6) + 3
p(6) = -216 + 36 + 30 + 3
p(6) = -147
Now we can calculate the displacement:
Displacement = Final position - Initial position
Displacement = -147 - 3
Displacement = -150
To find the distance traveled, we take the absolute value of the displacement:
Distance = |Displacement|
Distance = |-150|
Distance = 150
Therefore, during the first six seconds, the particle travels a distance of 150 units.