Acceleration from min/max derivatives

In summary: It would take the car about 0.5 seconds to go 100m if it started from rest, depending on how long you spend winding the flywheel.
  • #1
Turtlie
6
0

Homework Statement


x(t) = -0.01t^3 + t^2 - 20t + 4


Homework Equations


Min is when t = 12.3
Max is when t = 54.4


The Attempt at a Solution


I got -0.03t^2 + 2t - 20 as the derivative.
I substituted in t = 12.3 and 54.4 and got 0.02 and 0.19 which don't seem right at all.
Because: when t is 12.3, the position(x) is -100. When t is 54.4 the position(x) is 250.
 
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  • #2
Hello Turtlie :smile:

(try using the X2 icon just above the Reply box :wink:)
Turtlie said:
I got -0.03t^2 + 2t - 20 as the derivative.
I substituted in t = 12.3 and 54.4 and got 0.02 and 0.19 which don't seem right at all.

Well, to only 3 https://www.physicsforums.com/library.php?do=view_item&itemid=523" that looks acceptably near zero …

what happens if you solve that quadratic equation to 4 sig figs ? :smile:
 
Last edited by a moderator:
  • #3
Ah looks like I was right then, and thank you for the hint ;)

I've got another question.
A flywheel car is capable of giving a car a constant acceleration of 0.5m/s2, but only for as long as it is wound up for (if it is wound up for 5 seconds it will accelerate at 0.5m/s2. Once the flywheel is finished unwinding it will continue rolling at a constant velocity.

How long should you spend turning the flywheel if the length of the race is 100 meters. The time spent turning the flywheel counts towards the total time.

Honestly I'm completely stuck on this.
 
  • #4
Turtlie said:
Ah looks like I was right then, and thank you for the hint ;)

I've got another question.
A flywheel car is capable of giving a car a constant acceleration of 0.5m/s2, but only for as long as it is wound up for (if it is wound up for 5 seconds it will accelerate at 0.5m/s2. Once the flywheel is finished unwinding it will continue rolling at a constant velocity.

How long should you spend turning the flywheel if the length of the race is 100 meters. The time spent turning the flywheel counts towards the total time.

Honestly I'm completely stuck on this.

At a constant acceleration of 0.5m/s2, how long will it take the car to go 100m, if it starts from rest?
 
  • #5
SammyS said:
At a constant acceleration of 0.5m/s2, how long will it take the car to go 100m, if it starts from rest?

ah, but maybe it's better to spend less time winding it, and to let it cruise for the last 10 m or so? :smile:
 

What is acceleration from min/max derivatives?

Acceleration from min/max derivatives is a scientific concept that measures the rate of change of the velocity of an object over time. It is calculated by taking the second derivative of the position function of the object.

How is acceleration from min/max derivatives calculated?

To calculate acceleration from min/max derivatives, you must first determine the position function of the object. Then, you take the second derivative of that function to find the acceleration.

What is the difference between acceleration and velocity?

Acceleration is the rate of change of an object's velocity, while velocity is the rate of change of an object's position. In other words, velocity tells you how fast an object is moving, while acceleration tells you how quickly the object's velocity is changing.

Why is acceleration from min/max derivatives important?

Acceleration from min/max derivatives is important because it helps scientists and engineers understand and predict the motion of objects. It is also a key concept in many fields, such as physics, engineering, and astronomy.

Can acceleration from min/max derivatives be negative?

Yes, acceleration from min/max derivatives can be negative. This means that the object is decelerating, or slowing down, in its motion. A positive acceleration indicates that the object is accelerating, or speeding up, in its motion.

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