Maximizing Speed: Solving for the Maximum Speed of a Moving Particle

  • Thread starter Thread starter Minestra
  • Start date Start date
  • Tags Tags
    Maximum Speed
AI Thread Summary
The problem involves finding the maximum speed of a particle described by the function v(t) = ate^(-3t). The initial approach included taking the derivative and setting it to zero, but the user encountered issues because they incorrectly differentiated the function. The correct method requires applying the product rule to differentiate ate^(-3t). After realizing the mistake, the user expressed gratitude for the clarification. The discussion emphasizes the importance of proper differentiation techniques in solving calculus problems.
Minestra
Messages
15
Reaction score
0

Homework Statement


The instantaneous speed of a particle moving along one straight line is v(t) = ate-3t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

2. The attempt at a solution
As far as I know this is a max min type problem you'd see in calculus. So I take the derivative and set it to zero,
v'(t)= -3ae-3t
when I get to this point I kinda hit a wall because setting the problem to zero ultimately just gets me zero. I know the answer isn't zero. Any help with what I'm doing wrong will be much appreciated.
 
Physics news on Phys.org
Minestra said:

Homework Statement


The instantaneous speed of a particle moving along one straight line is v(t) = ate-3t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

2. The attempt at a solution
As far as I know this is a max min type problem you'd see in calculus. So I take the derivative and set it to zero,
v'(t)= -3ae-3t
when I get to this point I kinda hit a wall because setting the problem to zero ultimately just gets me zero. I know the answer isn't zero. Any help with what I'm doing wrong will be much appreciated.
You seem to have differentiated ae-3t, not ate-3t. Use the product rule.
 
You may want to check your derivative.
Sorry I'm late. See post #2.
 
haruspex said:
You seem to have differentiated ae-3t, not ate-3t. Use the product rule.
Hazah, thank you. Its a rather embarrassing mistake when I see it now.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top