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drek231 [11]
2 years ago
12

The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and

s(t) in metres.
a) When is the particle at rest?
b) When is the velocity positive?
c) Draw a diagram to illustrate the motion of the particle in the first 10 seconds.
d) Find the total distance traveled in the first 10 seconds.
Physics
1 answer:
8090 [49]2 years ago
4 0

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at 0.

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>
  • A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s

The velocity is positive at when the time of motion is as follows;

0.

The total distance traveled in the first 10 seconds is calculated as follows;

2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m

Learn more about motion of particles here: brainly.com/question/11066673

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(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

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where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

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The buoyant force is given by

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

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\rho_a = 1.29 kg/m^3 is the air density

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g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

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The mass of the helium gas inside the balloon is

m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

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F=B-W=4159 N-2635 N=1524 N

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(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

W'=(m'+m)g=B

where m' is the additional mass. Re-arranging the equation for m', we find

m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

B=\rho_a V g

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

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