Integrating Velocity: Distance = bt^3/3

AI Thread Summary
The discussion centers on a drag racing car's motion described by the velocity function v = bt^2. Participants question the validity of deriving distance from the given velocity, noting that distance and displacement may not be interchangeable without assumptions. The correct method to find distance involves integrating the velocity function, leading to the expression for distance traveled as bt^3/3. It is clarified that using the formula for constant velocity (v * time) is inappropriate here due to the variable nature of acceleration. Understanding integration is essential for solving this problem, as standard kinematic equations do not apply.
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Homework Statement



A drag racing car starts from rest at t = 0 and moves along a straight line with velocity given
by v = bt^2, where b is a constant. The expression for the distance traveled by this car from its
position at t = 0 is:

A. bt3
B. bt^3/3

Homework Equations



Velocity is change in position divided by change in time.

The Attempt at a Solution



Three questions:

1) Is this a bad question? I don't see how we can get distance traveled from what's given in this problem unless we make the assumption that distance = displacement in this problem.

2) If velocity is displacement / time, then why isn't the answer according to the problem v * time or bt^2 * t = bt^3?

3) Why is the integral of the velocity function in the question bt^3/3 - or different from simply velocity * time? Isn't the integral the area under the curve on a velocity time graph, or simply the y-axis * the x-axis (velocity * time)?
 
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You are correct; the answer is ##\frac{bt^3}{3}##. Try integrating ##bt^2##, treating ##b## as the constant, by calculus. Which rule do you need to use to integrate such an expression? This is one of the elementary derivative rules.
 
Qube said:
1) Is this a bad question? I don't see how we can get distance traveled from what's given in this problem unless we make the assumption that distance = displacement in this problem.
It does say the motion is in a straight line. The quantity bt2 is ≥0 or ≤0 throughout the motion depending on the sign of b.
2) If velocity is displacement / time, then why isn't the answer according to the problem v * time or bt^2 * t = bt^3?
This is only true if the body is moving at constant velocity. Is this the case here?
 
Qube said:
2) If velocity is displacement / time,
That's average velocity.
 
CAF123 said:
It does say the motion is in a straight line. The quantity bt2 is ≥0 or ≤0 throughout the motion depending on the sign of b.
This is only true if the body is moving at constant velocity. Is this the case here?

You are correct; I cannot use the usual kinematic equations here since they all assume that acceleration is constant, and taking the derivative of the velocity function (given) results in an expression with a variable.
 
If dx/dt = bt2, have you learned how to integrate this equation?
 
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