A particle moves along a line with a velocity v(t)=t2+3t−4, measured in meters per second. Find the total distance the particle travels from t=0 seconds to t=2 seconds.Enter an exact answer.Provide your answer below:

To find the total distance travelled by the particle, we need to find the integral of the absolute value of the velocity function from 0 to 2. However, since the velocity function is a polynomial, it may change sign, and we need to find where this happens in order to split the integral into parts.

First, let's find the roots of the velocity function v(t) = t^2 + 3t - 4. This is a quadratic equation, and we can solve it using the quadratic formula:

t = [-b ± sqrt(b^2 - 4ac)] / (2a)

For our equation, a = 1, b = 3, and c = -4. Plugging these values into the quadratic formula gives:

t = [-3 ± sqrt((3)^2 - 41(-4))] / (2*1) t = [-3 ± sqrt(9 + 16)] / 2 t = [-3 ± sqrt(25)] / 2 t = [-3 ± 5] / 2

This gives us two roots: t = 1 and t = -2. However, since we're only interested in the interval from t = 0 to t = 2, we only care about the root t = 1.

Now we split the integral of the velocity function into two parts: from 0 to 1, and from 1 to 2. We need to take the absolute value of each part, because distance is always positive.

The integral from 0 to 1 of v(t) dt is:

∫ from 0 to 1 (t^2 + 3t - 4) dt = [t^3/3 + 3t^2/2 - 4t] from 0 to 1 = 1/3 + 3/2 - 4 - (0) = -5/6

The integral from 1 to 2 of v(t) dt is:

∫ from 1 to 2 (t^2 + 3t - 4) dt = [t^3/3 + 3t^2/2 - 4t] from 1 to 2 = 8/3 + 6 - 8 - (1/3 + 3/2 - 4) = 3/2

The total distance travelled by the particle from t = 0 to t = 2 is the sum of these two distances:

|-5/6| + |3/2| = 5/6 + 3/2 = 13/6 meters.