The acceleration of a perticle as it moves along a straight line is geven by a=(2t-1) m/s2. If x=1m and v=2m/s when t=0s, determine the particle's velocity and position when t=6s. Also, determine the total distance the particle travels during this time period.
d^2x/dt^2 = 2 t - 1
dx/dt = t^2 - t + c
at t = 0, dx/dt = 2 so c = 2
so
v = dx/dt = t^2 - t + 2
x = (1/3) t^3 - (1/2) t^2 + 2 t + c
when t = 0 x = a so c = 1
x = (1/3) t^3 - (1/2) t^2 + 2 t + 1
so at t = 6
v = 36 -6 + 2 = 32 m/s
and
x = 67
total distance = 67-1 = 66
assuming x is always positive (sure looks that way
Thanks
To determine the velocity and position of the particle when t = 6s, we need to integrate the given acceleration function with respect to time.
Given: a = (2t - 1) m/s^2
Integrating a with respect to t will give us the particle's velocity function, v:
∫ a dt = ∫ (2t - 1) dt
Integrating, we get:
v = t^2 - t + C
We can find the value of C by using the initial conditions when t = 0s. Given: x = 1m and v = 2m/s.
When t = 0s, x = 1m:
x = ∫ v dt = ∫ (t^2 - t + C) dt
Integrating, we get:
x = (t^3/3 - t^2/2 + Ct) + D
Substituting the initial condition x = 1m and t = 0 into the equation, we can find D:
1 = (0^3/3 - 0^2/2 + C * 0) + D
1 = D
So, x = t^3/3 - t^2/2 + t + 1
Now, we can substitute t = 6s into the velocity and position equations to find the particle's velocity and position at t = 6s:
v = (6)^2 - 6 + C
v = 36 - 6 + C
v = 30 + C
x = (6^3/3 - 6^2/2 + 6 + 1)
x = 72 - 18 + 7
x = 61m
Therefore, the particle's velocity when t = 6s is 30 m/s, and its position is 61m.
To find the total distance traveled during this time period, we need to consider the signs of the velocity.
From t = 0s to t = 6s, the velocity function is: v = t^2 - t + C
For t = 0s to t = 3s, the velocity is positive (v > 0). This means the particle is moving in the positive direction.
For t = 3s to t = 6s, the velocity is negative (v < 0). This means the particle is moving in the negative direction.
Therefore, to find the total distance traveled, we need to calculate the sum of the distances traveled in the positive and negative directions.
For t = 0s to t = 3s:
Distance = ∫(t^2 - t + C) dt
Distance = (t^3/3 - t^2/2 + Ct) | from 0 to 3
Distance = (3^3/3 - 3^2/2 + C*3) - (0^3/3 - 0^2/2 + C*0)
Distance = 9 - 4.5 + 3C
For t = 3s to t = 6s:
Distance = ∫(t^2 - t + C) dt
Distance = (t^3/3 - t^2/2 + Ct) | from 3 to 6
Distance = (6^3/3 - 6^2/2 + C*6) - (3^3/3 - 3^2/2 + C*3)
Distance = 36 - 18 + 6C - 9 + 4.5 - 3C
Summing the distances in the positive and negative directions:
Total distance = Distance (0s to 3s) + Distance (3s to 6s)
Total distance = (9 - 4.5 + 3C) + (36 - 18 + 6C - 9 + 4.5 - 3C)
Total distance = 30 + 6C
Substituting the value of C (from above) into the equation, C = 30:
Total distance = 30 + 6(30)
Total distance = 30 + 180
Total distance = 210m
Therefore, the total distance traveled by the particle during this time period is 210m.
To find the velocity and position of the particle at t=6s, we will need to integrate the given acceleration function over the time interval from t=0s to t=6s.
Step 1: Calculate the integral of the given acceleration function with respect to time:
∫(2t-1) dt = t^2 - t + C
Step 2: Use the initial conditions (x=1m and v=2m/s at t=0s) to find the constant of integration, C.
When t=0s, we have x=1m and v=2m/s.
Plugging these values into the integrated equation, we get:
1 = (0)^2 - (0) + C
1 = C
So, the equation for velocity becomes:
v = t^2 - t + 1
Step 3: To find the particle's velocity (v) when t=6s, substitute t=6 into the velocity equation:
v = (6)^2 - (6) + 1
v = 36 - 6 + 1
v = 31 m/s
Therefore, the particle's velocity at t=6s is 31 m/s.
Step 4: To find the particle's position (x) when t=6s, integrate the velocity equation with respect to time:
∫(t^2 - t + 1) dt = (1/3)t^3 - (1/2)t^2 + t + C2
Step 5: Use the initial conditions (x=1m at t=0s) to find the constant of integration, C2.
When t=0s, we have x=1m.
Plugging these values into the integrated equation, we get:
1 = (1/3)(0)^3 - (1/2)(0)^2 + (0) + C2
1 = C2
So, the equation for position becomes:
x = (1/3)t^3 - (1/2)t^2 + t + 1
Step 6: To find the particle's position (x) when t=6s, substitute t=6 into the position equation:
x = (1/3)(6)^3 - (1/2)(6)^2 + (6) + 1
x = (1/3)(216) - (1/2)(36) + 6 + 1
x = 72 - 18 + 6 + 1
x = 61 m
Therefore, the particle's position at t=6s is 61 m.
Step 7: To determine the total distance traveled by the particle during the time interval from t=0s to t=6s, we need to consider both the negative and positive displacements.
When t<4, the acceleration is negative (2t-1<0), so the particle is decelerating. This means the displacement is negative.
For t>4, the acceleration is positive (2t-1>0), so the particle is accelerating. This means the displacement is positive.
To find the total distance traveled, we need to find the absolute values of the negative and positive displacements separately, and then add them together.
Negative displacement (d1) = | x(4) - x(0) |
= | [(1/3)(4)^3 - (1/2)(4)^2 + (4) + 1] - [(1/3)(0)^3 - (1/2)(0)^2 + (0) + 1] |
= | [64/3 - 8 + 4 + 1] - [0 - 0 + 0 + 1] |
= | [69/3] - [1] |
= | 23 - 1 |
= 22 m
Positive displacement (d2) = | x(6) - x(4) |
= | [(1/3)(6)^3 - (1/2)(6)^2 + (6) + 1] - [(1/3)(4)^3 - (1/2)(4)^2 + (4) + 1] |
= | [216/3 - 18 + 6 + 1] - [64/3 - 8 + 4 + 1] |
= | [80] - [29.67] |
= 50.33 m (rounded to two decimal places)
Total distance = d1 + d2
= 22 + 50.33
= 72.33 m (rounded to two decimal places)
Therefore, the total distance traveled by the particle during the time interval from t=0s to t=6s is 72.33 m.