Answer:
[tex]d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)[/tex]
Explanation:
We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.
The x-component of the length of the minute hand is:
[tex]d_{x}(t)=dcos(\theta (t))[/tex] (1)
Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:
[tex]\theta (t)=\omega *t[/tex] (2)
Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:
[tex]\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}[/tex]
Now, let's put this value on (2)
[tex]\theta (t)=\frac{\pi}{30}*t[/tex]
Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:
[tex]d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)[/tex]
I hope it helps you!
Answer:
Explanation:
The x-component of the length of the minute hand is:
[tex]d_x(t) = dcos(\theta(t))[/tex]
where,
d is the length of the minute hand [tex]d= \frac{D}{2}[/tex]
D is the diameter of the clock
t is the time (min)
using the angular kinematic equations,
[tex]\theta(t) = w * t[/tex]
the minute hand moves 90 degrees or 1/2 rad in 15 min, so,
[tex]w = \frac{\delta \theta}{\delta t}\\\\w = \frac{\theta_f - \theta_i}{t_f - t_i}\\\\w = \frac{\frac{\pi}{2} - 0}{15 - 0}\\\\w = \frac{\pi}{30}[/tex]
Therefore,
[tex]\theta(t) = \frac{\pi}{30} * t[/tex]
the length x(t) of the shadow of the minute hand as a function of time t,
[tex]d_x(t) = (\frac{D}{2})cos(\frac{\pi}{30} * t)[/tex]
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