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Anestetic [448]
3 years ago
7

A car is moving with speed 60 m/s and acceleration 4 m/s2 at a given instant. (a) using a second-degree taylor polynomial, estim

ate how far the car moves in the next second. m (b) would it be reasonable to use this polynomial to estimate the distance traveled during the next minute? yes no
Physics
1 answer:
pantera1 [17]3 years ago
6 0

Answer:

Explanation:

Using second degree taylor polynomials

let S(t) be position function and set S(0)=0

where S(0) is the initial position

Then v(t) = S^i(t) and a(t)=S^{ii}(t)

we have v(0) = 60m/s, a(0) = 4m/s^2

so T_{2}(t) =S(0)+v(0)t+\frac{a(0)}{2} t^2\\\\\\=0+60t+\frac{4}{2} t^2\\\\T_{2}(t)=60t+2t^2\\\\S(1)=T_{2}(1)=60+2=62m

b.) yes

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tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

M = 1.0 kg

r = 0.5 m

m = 10 g

I = 0.280 kg.m^2

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

I = I_{rim} + n * I_{spoke}

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I_{rim} = M*r^2 = 1.0 * 0.5^2

I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2

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0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )

Solving for n,

n = 36

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PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

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A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
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Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

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Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
2 years ago
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Answer:

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