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?