Displacement given the acceleration

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SUMMARY

The discussion focuses on calculating the distance traveled by a car accelerating at 3 m/s² until it reaches the same distance as a motorcycle moving at a constant speed of 20 m/s. The key kinematic equation to use is x(t) = x(0) + v(0)t + (1/2)at², where initial conditions for position and velocity must be specified. The initial displacement is assumed to be zero, while the initial velocity of the car is zero. The solution requires integrating acceleration to find velocity and subsequently position.

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  • Understanding of kinematic equations
  • Knowledge of differentiation and integration
  • Familiarity with constant acceleration concepts
  • Basic algebra skills for solving equations
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  • Study the derivation of kinematic equations for constant acceleration
  • Learn about initial conditions in physics problems
  • Explore the concept of integration in the context of motion
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Guillem_dlc
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Homework Statement


How could I get the distance traveled by a vehicle having only its constant acceleration as data?

Homework Equations


I know that the acceleration is the derivative of the velocity and the velocity is the derivative of the position.

The Attempt at a Solution


I don't know how to find the position because I have only the acceleration.
 
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use equations of kinematics
 
Guillem_dlc said:
having only its constant acceleration as data?
You can safely assume that the initial displacement is zero from the way the question is framed.

Unfortuately, you cannot justifiably make such an assumption about the initial velocity. This leave the problem without a solution.
 
At the instant when a traffic light turns green, a car starts to move with a constant acceleration of 3ms ^ 2, at that same moment, a motorcycle is placed at its speed at a constant speed of 20ms. The two vehicles continue advancing past the traffic light.

Calculate the distance, measured from the traffic light, which runs until it is at the height of the motorcycle

The speed of the car at the time it is at the height of the motorcycle
 
What you have just posted is not quite the problem in post #1.

Guillem_dlc said:
which runs until it is at the height of the motorcycle

As it stands, this does not quite make sense. Are you sure that height is the word you want heree?
 
Dr.D said:
What you have just posted is not quite the problem in post #1.
As it stands, this does not quite make sense. Are you sure that height is the word you want heree?
which runs until it is at the same distance of the motorcycle.

Thanks
 
Guillem_dlc said:

Homework Equations


I know that the acceleration is the derivative of the velocity and the velocity is the derivative of the position.

The Attempt at a Solution


I don't know how to find the position because I have only the acceleration

The opposite of differentiation is integration. You will need to think about the constants.
 
You need to specify the initial conditions of position and velocity to solve the problem. Then it is just an application of the 1D kinematic equation. x(t) = x(0) + v(0) t + a t^2 / 2.
 
Guillem_dlc said:
At the instant when a traffic light turns green, a car starts to move with a constant acceleration of 3ms ^ 2, at that same moment, a motorcycle is placed at its speed at a constant speed of 20ms. The two vehicles continue advancing past the traffic light.

Calculate the distance, measured from the traffic light, which runs until it is at the height of the motorcycle

The speed of the car at the time it is at the height of the motorcycle
Just to clean up the English a bit, I think you are saying that the car and motorcycle pass the traffic light at the same time, the car accelerating from rest at 3m/s2 and the motorcycle moving at a constant 20 m/s.
Suppose they meet again time t later. How far has each gone?
 
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