The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.
a(t) = 2t + 2, v(0) = −3, 0 ≤ t ≤ 4
(a) Find the velocity at time t.
(b) Find the distance traveled during the given time interval.
Derivative of displacement (with respect to time) is velocity.
Derivative of velocity (with respect to time) is acceleration.
Thus,
v(t) = ∫ (a(t)) dt
v(t) = ∫ (2t + 2) dt
v(t) = t^2 + 2t + C
It was said that at t = 0, v(0) = -3.
v(0) = (0)^2 + 2(0) + C = -3
v(0) = C = -3
Substitute,
v(t) = t^2 + 2t - 3
For the second question, evaluate v(t) at the interval from t=0 to t=4. Hope this helps~ `u`
I mean for the second question, integrate v(t) further and evaluate from t=0 to t=4:
D(t) = ∫ v(t)
D(t) = ∫ (t^2 + 2t - 3)
D(t) = (1/3)t^3 + t^2 - 3t + C
at t = 0:
D(0) = (1/3)(0)^3 + (0)^2 - 3(0) + C
D(0) = C
at t = 4:
D(4) = (1/3)(4)^3 + (4)^2 - 3(4) + C
D(4) = 76/3 + C
Thus,
76/3 + C - C = 76/3 meters
To find the velocity at time t, we can integrate the acceleration function with respect to t.
(a) Velocity can be found by integrating the acceleration function:
v(t) = ∫(2t + 2) dt
Applying the power rule of integration:
v(t) = t^2 + 2t + C
We need to find the constant C. Given v(0) = -3, we substitute t = 0 into the equation:
v(0) = (0)^2 + 2(0) + C
-3 = C
So, the velocity function is:
v(t) = t^2 + 2t - 3
(b) To find the distance traveled during the given time interval, we need to integrate the velocity function over the given interval.
The distance traveled is given by the definite integral:
distance = ∫[0, 4] |v(t)| dt
Since the velocity function can be positive or negative, we need to consider the absolute value.
Let's evaluate the integral:
distance = ∫[0, 4] |t^2 + 2t - 3| dt
Splitting the integral at the points where the expression inside the absolute value changes sign:
distance = ∫[0, 4] (t^2 + 2t - 3) dt + ∫[0, 4] -(t^2 + 2t - 3) dt
Integrating each term separately:
distance = ∫[0, 4] (t^2 + 2t - 3) dt - ∫[0, 4] (t^2 + 2t - 3) dt
Simplifying the integration:
distance = [t^3/3 + t^2 - 3t] [0, 4] - [t^3/3 + t^2 - 3t] [0, 4]
Substituting the limits of integration:
distance = [(4)^3/3 + (4)^2 - 3(4)] - [(0)^3/3 + (0)^2 - 3(0)]
Simplifying the expression:
distance = (64/3 + 16 - 12) - (0/3 + 0 - 0)
distance = 64/3 + 16 - 12
Final answer:
distance = 64/3 + 4 = 84/3 = 28 units
To find the velocity at time t, we can integrate the acceleration function a(t) with respect to time (t) and then add the initial velocity v(0).
(a) Find the velocity at time t:
Let's integrate the acceleration function with respect to t to find the velocity function v(t):
∫(a(t)) dt = ∫(2t + 2) dt
= t^2 + 2t + C
Since we know the initial velocity v(0) is -3, we plug in t = 0 to find the constant C:
v(0) = 0^2 + 2(0) + C = C
C = -3
Now, we have the velocity function v(t) = t^2 + 2t - 3.
(b) Find the distance traveled during the given time interval (0 ≤ t ≤ 4):
To find the distance traveled, we need to calculate the area under the velocity curve between t = 0 and t = 4.
Let's calculate the definite integral of the absolute value of the velocity function over the given time interval:
∫[0, 4] |v(t)| dt
Substituting the velocity function v(t) = t^2 + 2t - 3 into the integral:
∫[0, 4] |t^2 + 2t - 3| dt
This integral breaks into two separate cases, depending on the sign of the quantity inside the absolute value. So we need to find the roots of the equation t^2 + 2t - 3 = 0.
Solving t^2 + 2t - 3 = 0 by factoring or using the quadratic formula, we find t = -3 and t = 1 as the roots. Since the interval is [0, 4], we consider the interval [0, 1] and [1, 4] separately.
Let's calculate the distance traveled for each interval:
For the interval [0, 1]:
∫[0, 1] (t^2 + 2t - 3) dt
For the interval [1, 4]:
∫[1, 4] (t^2 + 2t - 3) dt
Evaluating these integrals will give us the distance traveled during each interval.