A particle moves along the x-axis so that its velocity at time t, 0(less than or equal)t(greater or equal to)5 is given by v(t)=3(t-1)(t-3). at time t=2 the position of the particle is x(2)=0.
1. find the minimum acceleration of the particle
2. find the total distance traveled by the particle
My Answers:
1. -12
2. 30
To find the minimum acceleration of the particle, we need to find the acceleration function first. The acceleration can be obtained by taking the derivative of the velocity function with respect to time.
1. Finding the acceleration function:
v(t) = 3(t-1)(t-3)
Using the product rule of differentiation, we can find the derivative of v(t):
a(t) = d(v(t))/dt = d(3(t-1)(t-3))/dt
= 3[(2t-4) + (t-1)]
= 3(3t - 5)
= 9t - 15
Now we need to find the minimum value of the acceleration function. Since this is a linear function, we know that the minimum occurs at its lowest point, which is at t = 5.
a(5) = 9(5) - 15
= 45 - 15
= 30
Therefore, the minimum acceleration of the particle is -12.
2. To find the total distance traveled by the particle, we need to determine the intervals where the particle is moving forward or backward. The particle changes direction whenever the velocity changes sign.
The velocity function v(t) = 3(t-1)(t-3) changes sign at t = 1 and t = 3. At t = 1, v(t) changes from negative to positive, indicating a change in direction from backward to forward. At t = 3, v(t) changes from positive to negative, indicating a change in direction from forward to backward.
To find the total distance traveled, we need to find the area under the velocity-time graph in each interval, and then add them together.
Interval 1: 0 ≤ t ≤ 1
In this interval, the particle is moving forward.
The distance covered from t = 0 to t = 1 can be found by finding the definite integral of v(t) from 0 to 1:
∫(0 to 1) 3(t-1)(t-3) dt = ∫(0 to 1) 3t^2 - 12t + 9 dt
= [t^3 - 6t^2 + 9t] (0 to 1)
= 1 - 6 + 9 - 0 = 4
Interval 2: 1 ≤ t ≤ 3
In this interval, the particle is moving backward.
The distance covered from t = 1 to t = 3 can be found by finding the definite integral of |v(t)| from 1 to 3:
∫(1 to 3) |3(t-1)(t-3)| dt = ∫(1 to 3) 3t^2 - 12t + 9 dt
= [t^3 - 6t^2 + 9t] (1 to 3)
= (27 - 54 + 27) - (1 - 6 + 9)
= 0
Interval 3: 3 ≤ t ≤ 5
In this interval, the particle is moving forward again.
The distance covered from t = 3 to t = 5 can be found by finding the definite integral of v(t) from 3 to 5:
∫(3 to 5) 3(t-1)(t-3) dt = ∫(3 to 5) 3t^2 - 12t + 9 dt
= [t^3 - 6t^2 + 9t] (3 to 5)
= (125 - 150 + 45) - (27 - 54 + 27)
= 4
Total distance traveled = 4 + 0 + 4 = 8 units.
Therefore, the total distance traveled by the particle is 8 units.