A cat walks in a straight line, which we shall call the x -axis with the positiv

Question

A cat walks in a straight line, which we shall call the x-axis with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and construct a graph of the feline's velocity as a function of time.

What distance does the cat move from t=0 to t=7.5s?

Answer: 26cm.

The solution can be found by computing displacement for t: [0,6] and t: [6,7.5] seperately.

My question is, when doing so, why do we use a negative acceleration for the first part of the graph (t: [0,6]) and then a positive acceleration for the second part (t: [6,7.5])?

                                Solution:
                            

                            

t: [0,6]    24.06 = 8*6+ 0.5(-1.33)(6^2)

t: [6,7.5]    1.5 = 0(1.5)+ 0.5(1.33)(1.5^2)

Accleration is still a constant negative. So why do we use a positive in the second equation? Using a negative acceleration would yield an answer of 24-1.5 = 22.5cm. But the answer is 24+1.5 = 26cm.

This is driving me insane. Please exlpain.

A cat walks in a straight line, which we shall call the x-axis with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and construct a graph of the feline's velocity as a function of time. What distance does the cat move from t=0 to t=7.5s? Answer: 26cm. The solution can be found by computing displacement for t: [0,6] and t: [6,7.5] seperately. My question is, when doing so, why do we use a negative acceleration for the first part of the graph (t: [0,6]) and then a positive acceleration for the second part (t: [6,7.5])? Solution: t: [0,6] 24.06 = 8*6+ 0.5(-1.33)(6^2) t: [6,7.5] 1.5 = 0(1.5)+ 0.5(1.33)(1.5^2) Accleration is still a constant negative. So why do we use a positive in the second equation? Using a negative acceleration would yield an answer of 24-1.5 = 22.5cm. But the answer is 24+1.5 = 26cm. This is driving me insane. Please exlpain.

Explanation / Answer

Very easy.

Slope = 8/6 = 4/3

distance covered in first 6 seconds = area under curve = 1/2 * 8*6 = 24 cm

distance covered in next 1.5 sec i.e. from 6 sec to 7.5 sec is area under the other part

Area = 1/2 * 1.5 * x

Similar triangles: x/1.5 = 8/6

=> x = 2

=> Area = 1/2 * 3/2 * 2 = 1.5cm

So the answer comes 25.5 = 26cm

Now coming to ur problem.

We use that equation for calculating the displacement and not the distance travelled

If u travel 2cm forward and 2 backward then net displacement is zero but distance is 2

Displacement = (+2) + (-2) = 0

But distance = |+2| + |-2| = 4

So in the eqaution for negative acceleration ur displacemenet comes out to be negative hence for distance purpose add it as positive quantity.

Please rate if it helps.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.