Find its position at the instants when velocity is zero

Click For Summary
SUMMARY

The discussion focuses on determining the position of a test car's front bumper at the instants when its velocity is zero. The position function is defined as x(t) = 2.11 m + (4.80 m/s²)t² - (0.100 m/s⁶)t⁶, while the velocity function is v(t) = 9.6t - 0.6t⁵. Participants clarified that to find the time when velocity equals zero, the equation must be factored to t(9.6 - 0.6t⁴) = 0, yielding t values of 0 and approximately 2.15 seconds. The acceleration at these instants is given by a(t) = 9.6 - 3t⁴.

PREREQUISITES
  • Understanding of polynomial functions and their derivatives
  • Familiarity with basic calculus concepts, particularly velocity and acceleration
  • Knowledge of factoring polynomials
  • Ability to interpret physical equations in the context of motion
NEXT STEPS
  • Study the process of finding critical points in polynomial functions
  • Learn how to apply the second derivative test for acceleration analysis
  • Explore the implications of zero velocity in kinematic equations
  • Investigate the relationship between position, velocity, and acceleration in motion analysis
USEFUL FOR

Physics students, engineers, and anyone interested in understanding motion dynamics, particularly in relation to polynomial equations and their applications in real-world scenarios.

AlexGM07
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
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=?)=??
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
961
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 3 ·
Replies
3
Views
913
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K