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 ≤ 5
(a) Find the velocity at time t.
v(t) = ______ m/s
(b) Find the distance traveled during the given time interval.
_____ m
Thank you!
To solve this problem, we need to integrate the acceleration function to find the velocity function, and then use the velocity function to find the distance traveled. Here's how we can do it:
(a) To find the velocity at time t, we need to integrate the acceleration function with respect to time. Let's call the integral constant C, which we'll determine using the initial velocity v(0).
∫(2t + 2)dt = t^2 + 2t + C
Since v(0) = -3, we can substitute t = 0 into the equation:
(0)^2 + 2(0) + C = -3
C = -3
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 integrate the absolute value of the velocity function over the interval 0 to 5:
∫[0 to 5] |t^2 + 2t - 3| dt
To evaluate this integral, we need to consider the sign changes of t^2 + 2t - 3 over the interval [0 to 5].
Let's find the critical points where t^2 + 2t - 3 = 0:
t^2 + 2t - 3 = (t + 3)(t - 1) = 0
t = -3 or t = 1.
This means that the function t^2 + 2t - 3 changes sign at t = -3 and t = 1.
Now we can split the integral into two parts and evaluate separately:
∫[0 to 5] (t^2 + 2t - 3) dt = ∫[0 to 1] (3 - t^2 - 2t) dt + ∫[1 to 5] (t^2 + 2t - 3) dt
Integrating each part:
∫[0 to 1] (3 - t^2 - 2t) dt = [3t - (1/3)t^3 - t^2/2] |[0 to 1]
= (3 - 1/3 - 1/2) - (0) = 1/6
∫[1 to 5] (t^2 + 2t - 3) dt = (1/3)t^3 + t^2 - 3t |[1 to 5]
= ((1/3)(5^3) + (5^2) - 3(5)) - ((1/3)(1^3) + (1^2) - 3(1))
= 75/3 + 25 - 15 - 1/3 - 1 + 3
= 93/3 = 31
Adding the two results together: 1/6 + 31 = 31 1/6
Therefore, the distance traveled during the given time interval is approximately 31 1/6 meters.
https://www.jiskha.com/display.cgi?id=1512879704
That was my bad, but he should have read and analyzed your solution.
The first part was correct and I got 155/3 m for B however, my homework system won't take that answer and doesn't want it as a decimal either. That is the answer, correct?
NO - I did this last night
The thing moves backwards from zero
then forward again
You found the displacement, not the distance.
v(t) = ∫a(t) dt = ∫(2t+2) dt = t^2+2t+C
v(0) = -3, so C = -3
v(t) = t^2+2t-3
s(t) = ∫[0,5] v(t) dt = ...