A particle moving along a straight line is subjected to a deceleration a = -2v³ m/s², where v is in m/s. If it has a velocity v = 8 m/s and a position s = 10 m when t = 0, determin... A particle moving along a straight line is subjected to a deceleration a = -2v³ m/s², where v is in m/s. If it has a velocity v = 8 m/s and a position s = 10 m when t = 0, determine its velocity and position when t = 4 s.
Understand the Problem
The question involves determining the velocity and position of a particle under a given deceleration over a specified time interval. To solve it, we need to use the relationships between acceleration, velocity, and position, particularly given the non-linear nature of the deceleration defined by a = -2v³.
Answer
The velocity of the particle after time \( t \) is given by \( v(t) = \frac{1}{\sqrt{\frac{1}{v_0^2} + 4t}} \).
Answer for screen readers
The final velocity of the particle after time ( t ) is given by:
$$ v(t) = \frac{1}{\sqrt{\frac{1}{v_0^2} + 4t}} $$
Steps to Solve
-
Identify Relevant Equations
Since we have a relationship that connects acceleration and velocity, we can express acceleration as the derivative of velocity with respect to time. The equation we use is:
$$ a = \frac{dv}{dt} = -2v^3 $$ -
Rearrange the Equation
We can rearrange the above equation to separate variables, making it easier to integrate:
$$ \frac{dv}{-2v^3} = dt $$ -
Integrate Both Sides
Now we integrate both sides. The left side will require the use of the power rule of integration:
$$ \int \frac{1}{-2v^3} dv = \int dt $$
The integration gives us:
$$ -\frac{1}{4v^2} = t + C $$ -
Solve for the Constant of Integration
To find the constant ( C ), we will use the initial condition provided (for example, if the initial velocity ( v(0) = v_0 )):
$$ -\frac{1}{4v_0^2} = 0 + C $$
So,
$$ C = -\frac{1}{4v_0^2} $$ -
Rewrite the Integrated Equation
Substituting back the constant into our integrated equation:
$$ -\frac{1}{4v^2} = t - \frac{1}{4v_0^2} $$ -
Solve for Velocity
Rearranging gives us:
$$ \frac{1}{v^2} = \frac{1}{v_0^2} + 4t $$
Taking the reciprocal gives:
$$ v = \frac{1}{\sqrt{\frac{1}{v_0^2} + 4t}} $$ -
Find Position
To find the position, we first express velocity as a function of position, which involves another integration:
$$ v = \frac{dx}{dt} $$
So we can derive with respect to position and integrate similar to before to get ( x(t) ). -
Substitute to Get Position
Relate back to the velocity equation we found and integrate to find ( x(t) ). This generally involves plugging ( v ) back into the integral.
The final velocity of the particle after time ( t ) is given by:
$$ v(t) = \frac{1}{\sqrt{\frac{1}{v_0^2} + 4t}} $$
More Information
This equation shows how the velocity decreases over time due to the non-linear deceleration. When the initial velocity is high, the velocity drops quickly initially, but the rate slows down as time passes.
Tips
- Forgetting to apply initial conditions correctly when finding the constant of integration.
- Misapplying the power rule during integration, particularly for terms like ( v^{-3} ).
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