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 ≤ 3
(a) Find the velocity at time t.
(b) Find the distance traveled during the given time interval.
To find the velocity at time t, we need to integrate the acceleration function with respect to time and then add the initial velocity v(0).
(a) Velocity at time t:
To find the velocity at time t, we need to integrate the acceleration function:
v(t) = ∫(a(t))dt
Given that a(t) = 2t + 2, we can integrate it to find v(t):
∫(2t + 2)dt = t^2 + 2t + C
Since we know the initial velocity v(0) = -3, we can substitute this value into the equation:
v(t) = t^2 + 2t + C
v(0) = 0^2 + 2(0) + C = C
Therefore, the velocity function is:
v(t) = t^2 + 2t - 3
(b) Distance traveled during the given time interval:
To find the distance traveled during the given time interval, we need to calculate the definite integral of the velocity function from t = 0 to t = 3:
Distance = ∫[0,3](v(t))dt
Using the velocity function v(t) = t^2 + 2t - 3, we can integrate it within the given time interval:
Distance = ∫[0,3](t^2 + 2t - 3)dt
This integral can be evaluated as follows:
Distance = [1/3 * t^3 + t^2 - 3t] from 0 to 3
= (1/3 * 3^3 + 3^2 - 3 * 3) - (1/3 * 0^3 + 0^2 - 3 * 0)
= (1/3 * 27 + 9 - 9) - (0)
= 9 + 9 - 0
= 18
Therefore, the distance traveled during the given time interval is 18 meters.
To find the velocity at time t, we need to integrate the acceleration function with respect to time.
(a)
Given: a(t) = 2t + 2, v(0) = -3
To find the velocity function v(t), we integrate a(t) with respect to t:
∫ a(t) dt = ∫ (2t + 2) dt
Using the power rule of integration, we can integrate each term separately:
∫ (2t + 2) dt = ∫ 2t dt + ∫ 2 dt
= t^2 + 2t + C
Adding the constant of integration (C), we get:
v(t) = t^2 + 2t + C
To find the constant C, we use the given initial velocity: v(0) = -3
Substitute t = 0 and v(0) = -3 into the equation:
v(0) = 0^2 + 2(0) + C
-3 = C
Therefore, the velocity function is:
v(t) = t^2 + 2t - 3
(b)
To find the distance traveled during the given time interval, we need to find the total displacement by integrating the velocity function over the given time interval [0, 3].
Using the definite integral:
∫[0,3] v(t) dt = ∫[0,3] (t^2 + 2t - 3) dt
Using the power rule of integration:
∫[0,3] (t^2 + 2t - 3) dt = [1/3 * t^3 + t^2 - 3t] [0,3]
Evaluating the integral at the upper and lower limits:
[1/3 * (3)^3 + (3)^2 - 3(3)] - [1/3 * (0)^3 + (0)^2 - 3(0)]
= [1/3 * 27 + 9 - 9] - 0
= 9 + 0 - 0
= 9
Therefore, the distance traveled during the given time interval is 9 meters.
if a(t) = 2t + 2
v(t) = t^2 + 2t + c
but v(0) = -3
-3 = 0 + 0 + c, c = 3
v(t) = t^2 + 2t + 3
s(t) = (1/3)t^3 + t^2 + 3t + k
s(0) = k
s(3) = 9 + 9 + 9 + k =27+k
distance travelled during the first 3 seconds = 27+k - k = 27 metres