Question

The velocity function is v(t) = - t^2 + 5 t - 6 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-3,5].

Answer 1

To find the displacement and distance traveled by the particle during the time interval [-3, 5], we need to integrate the absolute value of the velocity function over that interval.

The displacement is given by the definite integral of the velocity function:

Displacement = ∫[a, b] v(t) dt

In this case, a = -3 and b = 5, so the displacement is:

Displacement = ∫[-3, 5] (-t^2 + 5t - 6) dt

Integrating the velocity function:

Displacement = [- (1/3) t^3 + (5/2) t^2 - 6t] evaluated from -3 to 5

Displacement = [-(1/3)(5^3) + (5/2)(5^2) - 6(5)] - [-(1/3)(-3^3) + (5/2)(-3^2) - 6(-3)]

Simplifying the expression:

Displacement = [-(125/3) + (125/2) - 30] - [-(27/3) + (45/2) + 18]

Displacement = -125/6

Therefore, the displacement of the particle during the time interval [-3, 5] is -125/6 units.

To find the distance traveled, we need to integrate the absolute value of the velocity function:

Distance = ∫[a, b] |v(t)| dt

In this case, the velocity function is already given as v(t) = -t^2 + 5t - 6, so we don't need to take the absolute value.

Distance = ∫[-3, 5] (-t^2 + 5t - 6) dt

Using the same integral as before, we can find the distance traveled:

Distance = -125/6

Therefore, the distance traveled by the particle during the time interval [-3, 5] is also -125/6 units. Note that the negative sign indicates the direction of motion, not the actual distance.

Explanation:

no step by step explaination :(

Answer 2

Displacement = 58.7 in the negative direction

Distance = 59

Explanation:

Displacement

Displacement is the overall change in position from the initial to the final point, taking direction into account.

To find the displacement r(t) of the particle during the time interval [-3, 5] we integrate the expression for velocity v(t) with respect to time between the limits t = -3 and t = 5.

`\begin{aligned}r(t)=\displaystyle \int v(t)\; \text{d}t&=\int^5_(-3) \left(-t^2+5t-6\right)\; \text{d}t\\\\&=\left[-(t^(3))/(3)+(5t^2)/(2)-6t\right]^5_(-3)\\\\&=\left(-((5)^(3))/(3)+(5(5)^2)/(2)-6(5)\right)-\left(-((-3)^(3))/(3)+(5(-3)^2)/(2)-6(-3)\right)\\\\&=-(55)/(6)-(99)/(2)\\\\&=-(176)/(3)\end{aligned}`

Therefore, the displacement of the particle is 58.7 units in the negative direction.

`\hrulefill`

Distance

Distance is the total length traveled without considering direction. Therefore, to find the distance of the particle, we first need to determine if the particle changes direction within the given interval.

The instant(s) when the particle changes direction is the instant when the particle is momentarily at rest, meaning its velocity is zero.

Therefore, we can find the instant(s) when the particle changes direction by setting the velocity function v(t) to zero and solving for t:

`\begin{aligned}-t^2+5t-6&=0\\t^2-5t+6&=0\\t^2-3t-2t+6&=0\\t(t-3)-2(t-3)&=0\\(t-2)(t-3)&=0\\\\t-2&=0 \implies t=2\\t-3&=0 \implies t=3\end{aligned}`

So, the particle changes direction at t = 2 and t = 3.

Therefore, to find the total distance traveled, we need to evaluate the absolute values of the definite integrals of the velocity function over the intervals [-3, 2], [2, 3] and [3, 5], and then sum these absolute values.

`\displaystyle \textsf{Total distance}=\left|\int^2_(-3) v(t)\; \text{d}t\right|+\left|\int^3_(2) v(t)\; \text{d}t\right|+\left|\int^5_(3) v(t)\; \text{d}t\right|`

Evaluate each integral separately:

`\begin{aligned}\displaystyle \left|\int^2_(-3) \left(-t^2+5t-6\right)\; \text{d}t\right|&=\left[-(t^(3))/(3)+(5t^2)/(2)-6t\right]^2_(-3)\\\\&=\left|\left(-((2)^(3))/(3)+(5(2)^2)/(2)-6(2)\right)-\left(-((-3)^(3))/(3)+(5(-3)^2)/(2)-6(-3)\right)\right|\\\\&=\left|-(14)/(3)-(99)/(2)\right|\\\\&=\left|-(325)/(6)\right|\\\\&=(325)/(6)\end{aligned}`

`\begin{aligned}\displaystyle \left|\int^3_(2) \left(-t^2+5t-6\right)\; \text{d}t\right|&=\left[-(t^(3))/(3)+(5t^2)/(2)-6t\right]^3_(2)\\\\&=\left|\left(-((3)^(3))/(3)+(5(3)^2)/(2)-6(3)\right)-\left(-((2)^(3))/(3)+(5(2)^2)/(2)-6(2)\right)\right|\\\\&=\left|-(9)/(2)+(14)/(3)\right|\\\\&=\left|(1)/(6)\right|\\\\&=(1)/(6)\end{aligned}`

`\begin{aligned}\displaystyle \left|\int^5_(3) \left(-t^2+5t-6\right)\; \text{d}t\right|&=\left[-(t^(3))/(3)+(5t^2)/(2)-6t\right]^5_(3)\\\\&=\left|\left(-((5)^(3))/(3)+(5(5)^2)/(2)-6(5)\right)-\left(-((3)^(3))/(3)+(5(3)^2)/(2)-6(3)\right)\right|\\\\&=\left|-(55)/(6)+(9)/(2)\right|\\\\&=\left|-(14)/(3)\right|\\\\&=(14)/(3)\end{aligned}`

Now, sum the results to find the total distance traveled by the particle during the time interval [−3, 5]:

`\textsf{Total distance}=(325)/(6)+(1)/(6)+(14)/(3)=59`

Therefore, the distance travelled by the particle during the time interval [-3, 5] is 59 units.