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The position of a particle moving along the x axis is given in centimeters by x = 9.16 + 1.87 t3, where t is in seconds. Calculate (a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00 s; (c) the instantaneous velocity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s.

Respuesta :

Answer:

a)V=17.76 cm/s

b)V=22.44 cm/s

c)V=50.49 cm/s

d)V=35.06 cm/s

e)V=37.63 cm/s

Explanation:

Given that  

[tex]x=9.16+1.87t^3[/tex]

a)

When t= 2 s

[tex]x_1=9.16+1.87\times 2^3[/tex]

[tex]x_1=24.12\ cm[/tex]

When t= 3 s

[tex]x_2=9.16+1.87\times 3^3[/tex]

[tex]x_2=59.65\ cm[/tex]

So the average velocity V

[tex]V=\dfrac{x_2-x_1 }{t_2-t_1}[/tex]

[tex]V=\dfrac{59.65- 24.12}{3-1}[/tex]

V=17.76 cm/s

b)

When know that

[tex]V=\dfrac{dx }{dt}[/tex]

So

[tex]\dfrac{dx }{dt}=3\times 1.87t^2[/tex]

When t= 2 s

[tex]V=3\times 1.87\times 2^2[/tex]

V=22.44 cm/s

c)

At t= 3 s

[tex]V=3\times 1.87\times 3^2[/tex]

V=50.49 cm/s

d)

At t= 2.5 s

[tex]V=3\times 1.87\times 2.5^2[/tex]

V=35.06 cm/s

e)

At t= 2 s  ,x=24.12 cm

At t=3 s    x=59.65 cm

So midway of 24.12 cm and 59.65 cm is (24.12+59.65 )/2=41.88 cm

So when x=41.88 cm

[tex]x=9.16+1.87t^3[/tex]

[tex]41.88=9.16+1.87t^3[/tex]

t=2.59 s

So velocity

[tex]V=3\times 1.87t^2[/tex]

[tex]V=3\times 1.87\times 2.59^2[/tex]

V=37.63 cm/s

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