Find its position at the instants when velocity is zero

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Homework Help Overview

The discussion revolves around determining the position of a test car at the instants when its velocity is zero. The position and velocity functions are provided, and participants are exploring the relationship between these functions and the conditions under which velocity equals zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting the velocity function to zero to find the corresponding time values. There is uncertainty regarding the correct values obtained and the implications of those values for position and acceleration.

Discussion Status

Some participants have attempted to find the time when velocity is zero and have expressed confusion about the results. Others have pointed out the need to relate these times back to the position and acceleration functions, indicating a productive exploration of the problem.

Contextual Notes

There is a mention of the relationship between zero velocity and zero acceleration, suggesting that participants are considering the implications of these conditions on the problem at hand.

AlexGM07
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1.The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.11 m + (4.80 m/s^2)t^2 - (0.100 m/s^6)t^6.

x1 = ? (m)
x2 = ? (m)


2. v(t) = 9.6t - 0.6t^5



3. I tried setting v(t) = 0 and got -2, 0, 2. None of these turned out to be the right answer...I'm not sure what I'm doing wrong here.
 
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AlexGM07 said:
1.The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.11 m + (4.80 m/s^2)t^2 - (0.100 m/s^6)t^6.

x1 = ? (m)
x2 = ? (m)


2. v(t) = 9.6t - 0.6t^5



3. I tried setting v(t) = 0 and got -2, 0, 2. None of these turned out to be the right answer...I'm not sure what I'm doing wrong here.
If you take a look at the position formula x(t) = 2.11 m + (4.80 m/s^2)t^2 - (0.100 m/s^6)t^6, you can see that the position x depends on time t.
So, you have to find that position when velocity is equal to zero. That means that you have to find the moment t when velocity is equal to 0. Your velocity formula v(t) = 9.6t - 0.6t^5 is ok. Now you have to find moment t when velocity is 0.
v(t) = 9.6t - 0.6t^5
0=9.6t - 0.6t^5=t(9.6-0.6t^4)
I hope this helps.
 
method_man said:
If you take a look at the position formula x(t) = 2.11 m + (4.80 m/s^2)t^2 - (0.100 m/s^6)t^6, you can see that the position x depends on time t.
So, you have to find that position when velocity is equal to zero. That means that you have to find the moment t when velocity is equal to 0. Your velocity formula v(t) = 9.6t - 0.6t^5 is ok. Now you have to find moment t when velocity is 0.
v(t) = 9.6t - 0.6t^5
0=9.6t - 0.6t^5=t(9.6-0.6t^4)
I hope this helps.

Yeah, I figured that out this morning. I forgot to take the t out when I set the equation equal to zero...that was my big problem

But now I have to find its acceleration at the instants when the car has zero velocity

a(t) = 9.6 - 3t^4
But I'm not really sure what to do here. I can't take out a t, so how am I suppose to get two values of t? And when I get the values, what equation do I put them in: x(t), v(t), or a(t)?
 
AlexGM07 said:
But now I have to find its acceleration at the instants when the car has zero velocity
a(t) = 9.6 - 3t^4
When car has a zero velocity? Aren't those the same t's you have already calculated for x?
when the car has zero velocity
 
zero velocity→ zero acceleration
thus, a(t)=0=... ⇒t=? ⇒ x(t=?)=??
 

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