Question

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

Answer 1

Answer:

$V(t) = 2t^2 + 2t - 3$

$\Delta S = 46.667\ m$

Explanation:

We can find the velocity at time t using the formula:

$V(t) = V(0) + a*t$

Where V(t) is the velocity at time t, V(0) is the inicial velocity and a is the acceleration.

Then we have that:

$V(t) = -3 + (2t+2)*t$

$V(t) = 2t^2 + 2t - 3$

If we integrate the velocity in the time, we find the displacement (distance traveled):

$\int {V(t)} \, dt = 2t^3/3 + t^2 - 3t + c$

$\Delta S = \int\limits^4_0 {V(t)} \, dt = (2*4^3/3 + 4^2 - 3*4) - (2*0^3/3 + 0^2 - 3*0)$

$\Delta S = 46.667\ m$