(a) The velocity function v(t) is the integral of the acceleration function a(t). So, we integrate a(t) = 2t + 2 to get v(t) = t^2 + 2t + C. We know that v(0) = -15, so we can solve for C: -15 = 0 + 0 + C, so C = -15. Therefore, the velocity function is v(t) = t^2 + 2t - 15.

(b) The distance traveled is the integral of the absolute value of the velocity function from 0 to 5. We first need to find when the velocity is zero, as this is when the particle changes direction. Setting v(t) = 0 gives t^2 + 2t - 15 = 0. Factoring this gives (t-3)(t+5) = 0, so t = 3, -5. We discard -5 as it's outside our interval [0,5].

So, we integrate |v(t)| from 0 to 3 and from 3 to 5. This gives us the total distance traveled.

The integral from 0 to 3 of |v(t)| is ∫ from 0 to 3 of -(t^2 + 2t - 15) dt = [-1/3 t^3 - t^2 + 15t] from 0 to 3 = -9 + 9 + 45 = 45.

The integral from 3 to 5 of |v(t)| is ∫ from 3 to 5 of (t^2 + 2t - 15) dt = [1/3 t^3 + t^2 - 15t] from 3 to 5 = 50/3 + 25 - 75 - (9 + 9 - 45) = 50/3 - 10.

Adding these gives the total distance traveled: 45 + 50/3 - 10 = 35 + 50/3 = 105/3 + 100/3 = 205/3 = 68.33 meters.

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(a) The velocity function v(t) is the integral of the acceleration function a(t). So, we integrate a(t) = 2t + 2 to get v(t) = t^2 + 2t + C. We know that v(0) = -15, so we can solve for C: -15 = 0 + 0 + C, so C = -15. Therefore, the velocity function is v(t) = t^2 + 2t - 15.

(b) The distance traveled is the integral of the absolute value of the velocity function from 0 to 5. We first need to find when the velocity is zero, as this is when the particle changes direction. Setting v(t) = 0 gives t^2 + 2t - 15 = 0. Factoring this gives (t-3)(t+5) = 0, so t = 3, -5. We discard -5 as it's outside our interval [0,5].

So, we integrate |v(t)| from 0 to 3 and from 3 to 5. This gives us the total distance traveled.

The integral from 0 to 3 of |v(t)| is ∫ from 0 to 3 of -(t^2 + 2t - 15) dt = [-1/3 t^3 - t^2 + 15t] from 0 to 3 = -9 + 9 + 45 = 45.

The integral from 3 to 5 of |v(t)| is ∫ from 3 to 5 of (t^2 + 2t - 15) dt = [1/3 t^3 + t^2 - 15t] from 3 to 5 = 50/3 + 25 - 75 - (9 + 9 - 45) = 50/3 - 10.

Adding these gives the total distance traveled: 45 + 50/3 - 10 = 35 + 50/3 = 105/3 + 100/3 = 205/3 = 68.33 meters.

Knowee AI · Accepted Answer

(a) The velocity function v(t) is the integral of the acceleration function a(t). So, we integrate a(t) = 2t + 2 to get v(t) = t^2 + 2t + C. We know that v(0) = -15, so we can solve for C: -15 = 0 + 0 + C, so C = -15. Therefore, the velocity function is v(t) = t^2 + 2t - 15.

(b) The distance traveled is the integral of the absolute value of the velocity function from 0 to 5. We first need to find when the velocity is zero, as this is when the particle changes direction. Setting v(t) = 0 gives t^2 + 2t - 15 = 0. Factoring this gives (t-3)(t+5) = 0, so t = 3, -5. We discard -5 as it's outside our interval [0,5].

So, we integrate |v(t)| from 0 to 3 and from 3 to 5. This gives us the total distance traveled.

The integral from 0 to 3 of |v(t)| is ∫ from 0 to 3 of -(t^2 + 2t - 15) dt = [-1/3 t^3 - t^2 + 15t] from 0 to 3 = -9 + 9 + 45 = 45.

The integral from 3 to 5 of |v(t)| is ∫ from 3 to 5 of (t^2 + 2t - 15) dt = [1/3 t^3 + t^2 - 15t] from 3 to 5 = 50/3 + 25 - 75 - (9 + 9 - 45) = 50/3 - 10.

Adding these gives the total distance traveled: 45 + 50/3 - 10 = 35 + 50/3 = 105/3 + 100/3 = 205/3 = 68.33 meters.

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) = −15, 0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = t2+2t−15 m/s(b) Find the distance traveled during the given time interval.

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