The velocity function is v(t)=t^2−5t+4 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-1,5].

So I found the antiderivative of the function, which gave me t^3/3-5^2/2+4x+c. Then plugging in the given points and using the Fundamental Thm of Calculus, I got 6.

I know this is either the displacement or the distance traveled, but I'm not sure which. Can you help me understand which it is, and how to find the other value as well?

The value you found, 6, represents the displacement of the particle during the time interval [-1,5]. Displacement measures the change in position of the particle from its initial position to its final position. It can be positive or negative, depending on the direction of motion.

To find the distance traveled by the particle during the time interval, you need to consider both the magnitude and direction of its displacement. The distance traveled is always positive and represents the total path length covered by the particle.

To find the distance traveled, you can evaluate the definite integral of the absolute value of the velocity function over the given time interval [-1,5]. In this case, the velocity function is v(t) = t^2 - 5t + 4.

∫[-1,5] |v(t)| dt = ∫[-1,5] |t^2 - 5t + 4| dt

To evaluate this integral, you need to break it up into different intervals where the velocity function changes sign. In this case, the velocity function changes sign at t = 1 and t = 4. So, you can break the integral into two regions: [-1, 1] and [1, 5].

∫[-1,1] |v(t)| dt + ∫[1,5] |v(t)| dt

For the first region, [-1,1], the velocity function is negative. So, you can change the sign of the velocity function within the absolute value.

∫[-1,1] -(t^2 - 5t + 4) dt

For the second region, [1,5], the velocity function is positive, so you can keep the sign as it is.

∫[1,5] (t^2 - 5t + 4) dt

Evaluate the two integrals separately and take the sum of their absolute values to find the total distance traveled by the particle during the time interval [-1,5].

After evaluating these integrals, you should find that the total distance traveled by the particle during the time interval is 16 units.

Certainly! To understand whether 6 is the displacement or the distance traveled, let's first clarify what each term means in this context.

The displacement of an object refers to the change in its position from the initial point to the final point. It is a vector quantity that takes into account both the direction and magnitude of the change.

On the other hand, the distance traveled is a scalar quantity that represents the actual distance covered by the object, irrespective of the direction.

To determine the displacement, we need to find the change in position of the particle between the initial time and final time given.

In this case, we are given the velocity function, v(t) = t^2 - 5t + 4, which is the first derivative of the position function with respect to time. To find the position function, we need to integrate the velocity function.

By integrating v(t), we get the antiderivative of v(t) with respect to t, which is the position function, s(t):

s(t) = ∫ (t^2 - 5t + 4) dt

Simplifying this integral, we get:

s(t) = t^3/3 - 5t^2/2 + 4t + C

Now, to find the displacement between the initial and final times, we need to evaluate s(t) at those given points. In this case, the time interval is [-1, 5].

To find the displacement, we calculate:

Displacement = s(5) - s(-1)

Substituting the values of t into s(t), we get:

Displacement = (5^3)/3 - 5(5^2)/2 + 4(5) - ((-1)^3)/3 + 5((-1)^2)/2 - 4(-1)

Evaluating this expression gives us the displacement.

For the distance traveled, we need to consider the magnitude of the displacement regardless of its direction. Since the interval [-1, 5] includes both positive and negative displacements, we need to consider absolute values.

By taking the absolute value of the displacement, we can find the distance traveled:

Distance = |Displacement|

Now, let's calculate the actual values using the formulas above:

Displacement = (5^3)/3 - 5(5^2)/2 + 4(5) - ((-1)^3)/3 + 5((-1)^2)/2 - 4(-1)

Displacement = 125/3 - 125/2 + 20 + 1/3 + 5/2 + 4

Displacement = 6

Distance = |Displacement| = |6| = 6

Therefore, 6 is both the displacement and the distance traveled by the particle during the time interval [-1,5].

let x be distance

then dx/dt is velocity
so INT dx/dt dt has to be displacement

your antiderivative should be t^3/3-5t^2/2+4t+c.

The area under the curve of v(t) over time is displacement. If at any point that area is below the displacement =0 axis, then that "negative" area subtracts. For distance, you would want to add.
so you can integrate the abs value function of velocity to get distance. Or you can notice for the time -1 to zero, the function will be negative, figure that and add it to the other integral to get DISTANCE.

I did not check your displacement function to see if you got the right answer. Recheck it.