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Murljashka [212]

3 years ago

15

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

is x0 = 1.1 m at t0 = 0 s . part a at 2.0 s , what is the particle's position?

1 answer:

Nesterboy [21]3 years ago

5 0

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is <u>6.4m</u>

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A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i

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1 year ago

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3 years ago

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Calculate the time needed by the motorcycle to reach an acceleration of

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PART C)

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$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

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3 years ago

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Leokris [45]

Answer:

the answer the correct one is c

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3 years ago