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 thex-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 (the figure (Figure 1) ).

a)

Find the cat's velocity at t = 5.0 s .

Express your answer using two significant figures.

b)

Find the cat's velocity at t = 7.0 s .

Express your answer using two significant figures.

c)

What is the cat's acceleration at t=3.0s?

Express your answer using two significant figures.

d)

What is the cat's acceleration at t=6.0s?

Express your answer using two significant figures.

e)

What is the cat's acceleration at t=7.0s?

Express your answer using two significant figures.

f)

What distance does the cat move during the first 4.5 s?

Express your answer using two significant figures.

g)

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

Express your answer using two significant figures.

h)Sketch clear graph of the cat's acceleration as function of time, assuming that the cat started at the origin.

i)Sketch clear graph of the cat's position as function of time, assuming that the cat started at the origin.

"clear answers please"

Explanation / Answer

a) v is 8 cm/s at 0 sec .

and 0 cm/s at 6 s.

so equation of line,

v = 8 - (8/6)t

v = 8 - (4/3)t

at t= 5s

v = 8 - (4/3)5 = 1.33 cm/s

b) at t = 7s

v = 8 - (4*7/3) = - 1.33 cm/s

c) a = dv/dt = - 8/6 = - 1.33 cm/s^2

acceleration is constant.

a = - 1.33 cm/s^2

d) a = - 1.33 cm/s^2

e) a = - 1.33 cm/s^2

f) area under the v vs t curve give us distance traveled.

or d = integral of vdt = 8t - 2t^2/3

at t = 4.5 s

d = ( 8 x 4.5) - (2 x 4.5^2 / 3)

d = 22.5 cm

g) for 0 to 7.5s

first calculate distance for 0 to 6s,

d1 = (8 x 6) - (2 x 6^2 / 3) = 24 cm

for 6s to 7.5s

d2 = (8 (7.5 - 6 ) - (2 x (7.5^2 - 6^2) / 3)

d2 = 12 - 13.5 = 1.5 cm

total distance traveled = 24 + 1.5 = 25.5 cm

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