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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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Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

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\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

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