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. v(t) = m/s (b) Find the distance traveled during the given time interval. m
Question
Answer:
[tex]V(t) = 2t^2 + 2t – 3[/tex]
[tex]\Delta S = 46.667\ m[/tex]
Explanation:
We can find the velocity at time t using the formula:
[tex]V(t) = V(0) + a*t[/tex]
Where V(t) is the velocity at time t, V(0) is the inicial velocity and a is the acceleration.
Then we have that:
[tex]V(t) = -3 + (2t+2)*t[/tex]
[tex]V(t) = 2t^2 + 2t – 3[/tex]
If we integrate the velocity in the time, we find the displacement (distance traveled):
[tex]\int {V(t)} \, dt = 2t^3/3 + t^2 – 3t + c[/tex]
[tex]\Delta S = \int\limits^4_0 {V(t)} \, dt = (2*4^3/3 + 4^2 – 3*4) – (2*0^3/3 + 0^2 – 3*0)[/tex]
[tex]\Delta S = 46.667\ m[/tex]