How Do You Calculate Acceleration and Angular Speed in Physics Problems?

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Homework Help Overview

The discussion revolves around calculating acceleration and angular speed in physics problems, specifically focusing on kinematics and related rates. Participants are exploring how to determine constant acceleration for a vehicle decelerating over a distance and the angular speed of a radar dish tracking a vertically ascending shuttle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and integration to find acceleration and displacement. Questions arise about evaluating constants of integration and how to apply these concepts to the given problems. Some participants suggest using related rates to analyze the radar dish's angular speed.

Discussion Status

Several participants are actively engaging with the problems, providing insights and clarifications on the use of kinematic equations. There is an ongoing exploration of how to approach the second question regarding the radar dish, with some guidance offered on integrating acceleration and analyzing the geometry of the situation.

Contextual Notes

Participants note the importance of initial conditions and the definitions of displacement and velocity in their calculations. There is some confusion regarding the interpretation of displacement at time zero and its implications for the problems being discussed.

franz32
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1. What constant negative acceleration will enable a driver to decrease the speed from 120 km/hr to 60 km/hr while traveling a distance of 100m?

Where I got stuck [WIGS]: Let us say you don't know the formula used in kinematics... the acceleration is unknown. Let acceleration be a... so,
v = at + C(1), where C(1) is the first constant of integration and
s = 0.5at^2 + C(1)t + C(2) where C(2) is the second constant of integration.

Where will I evaluate my given values of velocity and distance? How can I
get the value of the acceleration?

2. A space shuttle climbs vertically with a constant acceleration of 10 yd/sec^2. If a radar-tracking dish, 1200 yd. from the shuttle's launch pad, follows the shuttle, how fast is the radar dish revolving 8 sec. after blast off?

[WIGS]: from the manual, the answer is 0.06 rad/sec. But what I don't
unerstand is how to arrive at that answer. To analyze the problem, it looks like a right triangle, your base is 1200 yd... how will antidifferentiation take place here? =)
 
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franz32 said:
1. What constant negative acceleration will enable a driver to decrease the speed from 120 km/hr to 60 km/hr while traveling a distance of 100m?

Where I got stuck [WIGS]: Let us say you don't know the formula used in kinematics... the acceleration is unknown. Let acceleration be a... so,
v = at + C(1), where C(1) is the first constant of integration and
s = 0.5at^2 + C(1)t + C(2) where C(2) is the second constant of integration.

Where will I evaluate my given values of velocity and distance? How can I
get the value of the acceleration?

2. A space shuttle climbs vertically with a constant acceleration of 10 yd/sec^2. If a radar-tracking dish, 1200 yd. from the shuttle's launch pad, follows the shuttle, how fast is the radar dish revolving 8 sec. after blast off?

[WIGS]: from the manual, the answer is 0.06 rad/sec. But what I don't
unerstand is how to arrive at that answer. To analyze the problem, it looks like a right triangle, your base is 1200 yd... how will antidifferentiation take place here? =)

The trick here is that you can evaluate the constants of integration by letting t equal zero.

So, starting from the definition of acceleration:

[tex]a = dv/dt[/tex]

[tex]a~dt = dv[/tex]

[tex]\int a~dt = \int dv[/tex]

[tex]at + C_1 = v[/tex]

If t = 0, then [itex]C_1 = v[/itex]. The value of v when t = 0 must be [itex]v_0[/tex] by definition (the "initial" velocity), so:<br /> <br /> [tex]at + v_0 = v[/tex]<br /> <br /> Using the definition of velocity and integrating again:<br /> <br /> [tex]at + v_0 = dx/dt[/tex]<br /> <br /> [tex]at~dt + v_0~dt = dx[/tex]<br /> <br /> [tex]\int at~dt + \int v_0~dt = \int dx[/tex]<br /> <br /> [tex]\frac {at^2}{2} + v_0 t + C_2 = x[/tex]<br /> <br /> We use the same trick as before - at t = 0, [itex]C_2 = x[/itex], and we know that, by definition, [itex]x = x_0[/itex] at t = 0 (the initial position), so [itex]C_2 = x_0[/itex] and:<br /> <br /> [tex]\frac {at^2}{2} + v_0 t = x - x_0[/tex]<br /> <br /> To eliminate t, we take the equation we got after the first integration, solve for t, substitute into the most recent equation, simplify, and rearrange the result:<br /> <br /> [tex]v = v_0 + at[/tex]<br /> <br /> [tex]\frac {v - v_0}{a} = t[/tex]<br /> <br /> [tex]\frac {a}{2} \cdot \frac{v^2 - 2vv_0 + v_0^2}{a^2} + \frac {vv_0 - v_0^2}{a} = x - x_0[/tex]<br /> <br /> Multiply through by 2a and simplify:<br /> <br /> [tex]v^2 - v_0^2 = 2a(x - x_0)[/tex]<br /> <br /> [tex]\frac {v^2 - v_0^2}{2(x - x_0)} = a[/tex] <br /> <br /> Hopefully you don't have to prove your kinematics equations on every problem <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /> <br /> <br /> I haven't tried the second question yet, but it sounds like you can integrate the acceleration of the shuttle to find out how quickly that side of the triangle is growing with respect to time, then treat it as a related rates problem.[/itex]
 
1) You know that v(0) = 120 km/h, and s(0) = 0.

2) Since you know the acceleration, you can calculate (with integration) how far the spaceship has flown at the time t (call this s(t)). Let alpha be the angle between the radar dish and the shuttle, then tan(alpha) = s(t)/1200. Then you can calculate how fast alpha changes at t = 8, using the derivative...

*edit* Oops, I see someone's already answered. That's what I get for taking so long when typing up my answer :P
 
Zorodius said:
[tex]We use the same trick as before - at t = 0, [itex]C_2 = x[/itex], and we know that, by definition, [itex]x = x_0[/itex] at t = 0 (the initial position), so [itex]C_2 = x_0[/itex] and:<br /> <br /> Um, by looking to your final "equation", I noticed that x - x(0) is equal to 0 according to the quote given above... I hope you can clarify that to me, by the way, thanks for your clues. =)[/tex]
 
Yes, at t = 0, the change in displacement (x - x0) is zero - the driver hasn't had a chance to move yet, since no time has passed. x is the current displacement, and x0 is the initial displacement. At any other time, x - x0 is not zero.

The problem says that the driver has traveled through a distance of 100m. We don't immediately know how long it took him to do that, but it doesn't matter - we know x0 = 0, x = 100 m, and so x - x0 = 100 m.
 
Ah, I see. =)
Thanks.
 

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