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vfiekz [6]
3 years ago
14

(b) Find the position, velocity, and acceleration of the mass at time t = 5π/6.

Physics
1 answer:
Juli2301 [7.4K]3 years ago
4 0
Solution 
x(t) = 8 cos t, x(5π/6)= 8 cos(<span>5π/6)
</span>cos(5π/6)=cos(3π/6 + 2π/6 )=cos(π/3 +π/2)= - sin π/3  (cos (x+<span>π/2)= -sinx)
</span>x(t) = -8sin <span>π/3 = - 4 .sqrt3
</span>v(t) = -8sint = -8sin (π/3 +<span>π/2)= -8 cosπ/3 </span>(sin (x+π/2)= cosx)
v(t) =<span> -8 cosπ/3 = -8/2= - 4 
</span>a(5π/6) = - 8cost = -(- sin π/3)= 4 .<span>sqrt3
</span>a(5π/6) =  4 .<span>sqrt3</span>
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A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

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y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

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