# Integrating velocity

1. Jan 19, 2014

### Qube

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Velocity is change in position divided by change in time.

3. 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)?

2. Jan 19, 2014

### NasuSama

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.

3. Jan 19, 2014

### CAF123

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?

4. Jan 19, 2014

### haruspex

That's average velocity.

5. Jan 19, 2014

### Qube

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.

6. Jan 19, 2014

### Staff: Mentor

If dx/dt = bt2, have you learned how to integrate this equation?