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):

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

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

$\Delta S = 46.667\ m$

Answer 2

We want to get the motion equations for a particle, given that we know its acceleration and its initial velocity.

a) $v(t) = (1m/s^3)*t^2 + (2m/s^2)*t - 3m/s$

b) the distance traveled in that interval is 25.33 meters.

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Let's see how to et the solutions:

a) We want to find the velocity equation, to get it, we can just integrate the acceleration equation such that the constant of integration will be the initial velocity.

The acceleration equation is:

$a(t) = (2m/s^3)*t + 2m/s^2$

To get the velocity we integrate:

$v(t) = (1/2)*(2m/s^3)*t^2 + (2m/s^2)*t + v_0$

Where v₀ is the initial velocity, which we know is -3m/s

Then the velocity equation is:

$v(t) = (1m/s^3)*t^2 + (2m/s^2)*t - 3m/s$

b) Now we want to find the distance traveled in the interval 0s ≤ t ≤ 4s

Then first we need to get the position equation, to do it we need to integrate the velocity equation.

$p(t) = (1/3)*(1m/s^3)*t^3 + (1/2)*(2 m/s^2)*t^2 - (3m/s)*t$

Where the constant of integration is ignored because we want to compute a difference.

The distance traveled in that interval is given by:

$p(4s) - p(0s) = p(4s) = (1/3)*(1m/s^3)*(4s)^3 + (1/2)*(2 m/s^2)*(4s)^2 - (3m/s)*4s = 25.33m$

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