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miskamm [114]
3 years ago
8

The position of an object is given by the equation x = 4.0t2 - 2.0 t - 4.5where x is in meters and t is in seconds. What is the

instantaneous acceleration, velocity and position of the object at t = 3.0 s? At what time t is the velocity zero. At what time t is the position zero. At what time t is the velocity zero.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

j

Explanation:

x = 4 t^2 - 2 t - 4.5

Position at t = 3 s

x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m

Velocity at t = 3 s

v = dx / dt = 8 t - 2

v ( t = 3 s) = 8 x 3 - 2 = 22 m/s

Acceleration at t = 3 s

a = dv / dt = 8

a ( t = 3 s ) = 8 m/s^2

When is the velocity = 0

v = 0

8 t - 2 = 0

t = 0.25 second

When is the position = 0

x = 0

4 t^2 - 2 t - 4.5 = 0

t = \frac{2 \pm \sqrt{4 + 72}}{8}

t = 1.4 second

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lukranit [14]

Answer:

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b)

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a)

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c)

For constant resistance and 110 V the power becomes:

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d)

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

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So, according to the law of conservation of energy

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Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

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Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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