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sergejj [24]
3 years ago
11

Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the

seven seconds that I’m watching it (so from t = 0 to t = 7). (a) What is the velocity function, v(t), for the motion of the squirrel? (b) What is the acceleration function, a(t), for the motion of the squirrel? (c) At the four second mark, is the squirrel moving up the tree or down the tree? Justify your answer.
Physics
1 answer:
marin [14]3 years ago
6 0

Answer:

Explanation:

Given

Position of squirrel is given by

s(t)=t^3-12t^2+36t

Velocity is given by

v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}

v(t)=3t^2-12\times 2t+36

v(t)=3t^2-24t+36

(b) acceleration is given by

a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}

a(t)=6t-24

(c)at s(3)=3^3-12(3)^2+36(3)

s(3)=27\ m

at s(4)=4^3-12(4)^2+36(4)

s(4)=16\ m

at t=3\ s Position is 27\ m and at t=4\ s position is 16\ m

therefore squirrel is moving down

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Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

∴

(a) 816 kg × v = 2650 kg × 16.0 km/h

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v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

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3 years ago
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3 0
3 years ago
determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund
wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is a_G= 6.78 m/s^2

<u>Explanation</u>:

N_A+N_B-Mg = 0

-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0

0.8N_B +0.8N_A = 975a_G

N_A+N_B = 9564.75 -------------(1)

-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0

-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0

-2.26N_A -0.06N_B= 0 ----------------(2)

Solving the equation (1) and(2)

N_A + N_B = 9564.75

-2.26N_A-0.06N_B=0

N_A = -260.85N

N_B = 9825.60N

\mu_s N_B + \mu_s N_A = 975a_G

0.8(9825.60)+0.8(-260.85) = 975a_Ga_G=\frac{7651.8}{975}a_G_1=7.4848m/s^2

Next lets assume that the front wheels contact with the ground N_A = 0

F_B = Ma_G

N_B = M_g

N_B - M_g = 0

N_B(b-a) –F_Bh = 0

F_B = 975a_G

N_B-975(9.8) = 0

N_B=9564.75N

9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

F_B = 6608.3

F_B = Ma_G

6608.3 = 975a_G

a_G = 6.7778 m/s^2

a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

3 0
3 years ago
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