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 function is the derivative of the velocity function.
1. To find the acceleration function, we differentiate the velocity function v(t)=3(t-1)(t-3) with respect to t.
v'(t) = d/dt [3(t-1)(t-3)]
= 3[(t-1)(2t-6) + (t-3)(2)]
= 3[(2t^2 - 8t + 6) + (t - 3)]
= 3(2t^2 - 7t + 3)
So, the acceleration function a(t) = 3(2t^2 - 7t + 3).
To find the minimum acceleration, we need to find the critical points of the acceleration function. These are the values of t for which the derivative of the acceleration function equals zero.
2. Setting a'(t) = d/dt [3(2t^2 - 7t + 3)] equal to zero:
a'(t) = 0
6t - 7 = 0
6t = 7
t = 7/6
So, the critical point is at t = 7/6.
To determine whether it is a minimum or maximum point, we can examine the concavity of the acceleration function.
3. To find the concavity of the acceleration function, we differentiate the acceleration function a(t) with respect to t.
a''(t) = d/dt [3(2t^2 - 7t + 3)]
= 6t - 7
4. Substitute the critical point t = 7/6 into the second derivative:
a''(7/6) = 6(7/6) - 7
= 7 - 7
= 0
Since the second derivative is equal to zero at t = 7/6, it means that the acceleration function changes concavity at this point.
Therefore, the critical point t = 7/6 is a point of inflection, not a minimum or maximum point. We cannot determine the minimum acceleration based on the given information.
For the second question:
To find the total distance traveled by the particle, we need to integrate the absolute value of the velocity function over the time interval [0, 5]. This will give us the total distance traveled.
5. Integrating the absolute value of the velocity function v(t) = 3(t-1)(t-3) with respect to t over the interval [0, 5]:
∫[0,5] |v(t)| dt
= ∫[0,5] |3(t-1)(t-3)| dt
To find the total distance, we need to consider the two separate intervals where the velocity function is positive and negative.
On the interval [0,1], (t-1)(t-3) is positive, so we can drop the absolute value signs:
∫[0,1] 3(t-1)(t-3) dt
= 3 ∫[(t^2 - 4t + 3)] dt
= 3 [ (1/3)t^3 - 2t^2 + 3t ] | [0,1]
= 3 ([1/3] - 2 + 3) - 3([0] - 0 + 0)
= 3 (4/3)
= 4
On the interval [1,5], (t-1)(t-3) is negative, so we need to consider the absolute value:
∫[1,5] 3(t-1)(t-3) dt
= 3 ∫[(3t^2 - 12t + 9) - (t^2 - 4t + 3)] dt
= 3 [ (2/3)t^3 - 4t^2 + 6t ] | [1,5]
= 3 ([(2/3)(125) - 4(25) + 6(5)] - [(2/3)(1) - 4(1) + 6(1)])
= 3 (100/3 - 100/3)
= 0
The total distance traveled is the sum of the distances traveled on each interval:
Total distance = 4 + 0
= 4
Therefore, the total distance traveled by the particle is 4 units.
To find the minimum acceleration of the particle, we can differentiate the velocity function with respect to time, t.
1. Differentiate v(t) = 3(t-1)(t-3) with respect to t:
v'(t) = 6t - 18
2. The acceleration of the particle is given by the derivative of the velocity function. So, a(t) = v'(t).
3. At t = 2, substitute t = 2 into v'(t):
a(2) = v'(2)
a(2) = 6(2) - 18
a(2) = 12 - 18
a(2) = -6
Therefore, the minimum acceleration of the particle is -6.
To find the total distance traveled by the particle, we need to find the area under the velocity curve between t = 0 and t = 5.
1. Integrate v(t) = 3(t-1)(t-3) with respect to t:
∫ v(t) dt = ∫ 3(t-1)(t-3) dt
2. Expand and simplify the integrand:
∫ 3(t^2 - 4t + 3) dt
∫ 3t^2 - 12t + 9 dt
3. Integrate term by term:
∫ 3t^2 dt - ∫ 12t dt + ∫ 9 dt
t^3 - 6t^2 + 9t + C
4. Evaluate the definite integral from t = 0 to t = 5:
∫[0,5] 3t^2 - 12t + 9 dt
[t^3 - 6t^2 + 9t] from 0 to 5
(5^3 - 6(5^2) + 9(5)) - (0^3 - 6(0^2) + 9(0))
(125 - 150 + 45) - (0 - 0 + 0)
20
Therefore, the total distance traveled by the particle is 20 units.