the acceleration of a particle as it moves along a straight line is given by a=(2t-1)m/s^2, where t is in seconds.if s=1m and v= 2m/s when t=0,determine the particles velocity and position when t=6s. Also, determine the total distance the particle travels during this time period.

a = 2t - 1

then v = t2 - t + c
when t = 0, v = 2
2 = c
then v = t^2 - t + 2

s = (1/3)t^3 - (1/2)t^2 + 2t + k
when t = 0 , s = 1
1 = k

s = (1/3)t^3 + (1/2)t^2 + 2t + 1

plug in t=6 to get your results.

To determine the particles velocity and position when t = 6s, we need to integrate the given acceleration function.

Step 1: Integrate the acceleration function to find the particles velocity:
∫(2t-1) dt = t^2 - t + C

Given that v = 2 m/s when t = 0, we can solve for C:
v = t^2 - t + C
Substituting t = 0 and v = 2:
2 = 0^2 - 0 + C
C = 2

So, the equation for the particles velocity is:
v = t^2 - t + 2

Step 2: Integrate the velocity function to find the particles position:
∫(t^2 - t + 2) dt = (1/3) t^3 - (1/2) t^2 + 2t + D

Given that s = 1 m when t = 0, we can solve for D:
s = (1/3) t^3 - (1/2) t^2 + 2t + D
Substituting t = 0 and s = 1:
1 = 0 - 0 + 0 + D
D = 1

So, the equation for the particles position is:
s = (1/3) t^3 - (1/2) t^2 + 2t + 1

Step 3: Substitute t = 6 into the equations for v and s to find the particles velocity and position when t = 6s:
v = (6)^2 - 6 + 2
v = 36 - 6 + 2
v = 32 m/s

s = (1/3) (6)^3 - (1/2) (6)^2 + 2(6) + 1
s = 72 - 18 + 12 + 1
s = 67 m

Therefore, when t = 6s, the particles velocity is 32 m/s and its position is 67 m.

Step 4: Calculate the total distance traveled by the particle during this time period.
To find the total distance, we need to consider both the positive and negative displacements along the path.

The particle travels a total distance equal to the absolute value of the displacement.

Displacement s(6) = 67 m (found in Step 3)
Displacement s(0) = 1 m (given in the problem)

As the particle moves in a straight line, we can calculate the total distance traveled using the formula:
Total distance traveled = |s(6) - s(0)|

Total distance traveled = |67 - 1|
Total distance traveled = 66 m

Therefore, the particle travels a total distance of 66 meters during the time period t = 0 to t = 6.

To determine the velocity and position of the particle when t=6s, we need to integrate the given acceleration equation.

Let's first integrate the acceleration equation to find the velocity equation.
∫a dt = ∫(2t - 1) dt
v = ∫(2t - 1) dt
v = t^2 - t + C

To find the constant C, we use the initial condition where t = 0 and v = 2 m/s.
2 = (0)^2 - (0) + C
C = 2

So, the velocity equation is v = t^2 - t + 2 m/s.

Now, we integrate the velocity equation to find the position equation.
∫v dt = ∫(t^2 - t + 2) dt
s = (t^3/3) - (t^2/2) + 2t + D

To find the constant D, we use the initial condition where t = 0 and s = 1 m.
1 = (0^3/3) - (0^2/2) + 2(0) + D
D = 1

So, the position equation is s = (t^3/3) - (t^2/2) + 2t + 1 m.

To find the particle's velocity when t = 6s, substitute t = 6 into the velocity equation:
v = (6)^2 - (6) + 2
v = 36 - 6 + 2
v = 32 m/s

To find the particle's position when t = 6s, substitute t = 6 into the position equation:
s = (6^3/3) - (6^2/2) + 2(6) + 1
s = 72 - 18 + 12 + 1
s = 67 m

To find the total distance the particle travels during this time period, we need to calculate the distance traveled from t = 0 to t = 6s. The distance traveled is the magnitude of the change in position.

The distance traveled = |s(6) - s(0)|
= |67 - 1|
= 66 m

Therefore, when t = 6s, the particle's velocity is 32 m/s, its position is 67 m, and it has traveled a total distance of 66 m during this time period.