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a cart is constrained to move along a straight line. a varying net force along the direction of motion is exerted on the cart. the cart’s velocity v as a function of time t is shown in the graph above. the five labeled points divide the graph into four sections. for which segment does the cart move the greatest distance?

Sagot :

MrRoyal

The distance between two points is the number of units between them.

The cart move the greatest distance in segment AB

Calculate distance AB

To do this, we calculate the perimeter of the quarter of the ellipse

This is calculated using:

[tex]\mathbf{P = \frac{\pi}{4} (a + b) ( 3 \times \frac{(a b)^2}{(a+b)^2(\sqrt{-3\frac{(ab)^2}{a + b)^2 }+ 4}\ + 10} + 1}) }[/tex]

Where:

[tex]\mathbf{a = 4ms^{-1}}[/tex]

[tex]\mathbf{b = 0.6s}[/tex]

Using a calculator,

[tex]\mathbf{P = 4.13m}[/tex]

So:

[tex]\mathbf{AB = 4.13m}[/tex]

Calculate distance BC

From the graph (see attachment), we have:

[tex]\mathbf{time = 0.2s}[/tex]

[tex]\mathbf{velocity = 8ms^{-1}}[/tex]

So, the distance traveled is:

[tex]\mathbf{distance = time \times velocity}[/tex]

[tex]\mathbf{distance = 0.2s \times 8ms^{-1}}[/tex]

[tex]\mathbf{distance = 1.6m}[/tex]

So:

[tex]\mathbf{BC = 1.6m}[/tex]

Calculate distance CD

From the graph, we have:

[tex]\mathbf{u = 8}[/tex] --- initial velocity at point C

[tex]\mathbf{v = 0}[/tex] ---- final velocity at point D

[tex]\mathbf{t = 0.2}[/tex] --- time of travel

So, the acceleration is:

[tex]\mathbf{a = \frac{v - u}{t}}[/tex]

[tex]\mathbf{a = \frac{0 - 8}{0.2} = -40ms^{-2}}[/tex]

The distance travelled is:

[tex]\mathbf{s = ut + \frac{1}{2}at^2}[/tex]

So, we have:

[tex]\mathbf{s = 8 \times 0.2 - \frac{1}{2} \times 40 \times 0.2^2}[/tex]

[tex]\mathbf{s = 1.6 - 0.8}[/tex]

[tex]\mathbf{s = 0.8}[/tex]

So:

[tex]\mathbf{CD = 0.8m}[/tex]

Calculate distance DE

From the graph, we have:

[tex]\mathbf{u = 0}[/tex] --- initial velocity at point D

[tex]\mathbf{v = -4}[/tex] ---- final velocity at point E

[tex]\mathbf{t = 0.6}[/tex] --- time of travel

So, the acceleration is:

[tex]\mathbf{a = \frac{v - u}{t}}[/tex]

[tex]\mathbf{a = \frac{-4 - 0}{0.6} = -6.67ms^{-2}}[/tex]

The distance travelled is:

[tex]\mathbf{s = ut + \frac{1}{2}at^2}[/tex]

So, we have:

[tex]\mathbf{s = 0 \times 0.6 - \frac{1}{2} \times 6.67 \times 0.6^2}[/tex]

[tex]\mathbf{s = 0 - 1.2}[/tex]

[tex]\mathbf{s = - 1.2}[/tex]

So:

[tex]\mathbf{DE = 1.2m}[/tex]

By comparing the distances, segment AB is the longest i.e. 4.13 m

Hence, the cart move the greatest distance in segment AB

Read more about distance and rates at:

https://brainly.com/question/21149231

View image MrRoyal
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