Hello again!
Why is it D and not A ? The passage says "decrease linearly with velocity".
Thanks!
C/P #8

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Re: C/P #8
Hi ellie 
This is a tough question, b/c it's a great example of how the MCAT doesn't technically test calculus, but it tests the ideas underlying calculus. The passage says that force decreases linearly with speed, which means that acceleration decreases linearly with speed. But the question is asking us to look at a graph of velocity. Keeping in mind that acceleration is the change in velocity over time, or the slope on a graph of V vs. t, we need to look for the graph that has a slope that changes linearly. In A and B, the graph is linear, but the slope is constant, so that's not what we're looking for. C and D have a slope that changes as a linear function of time.
The above explanation doesn't require calculus, but if you have taken calculus, what this boils down to is that you integrate acceleration to get velocity, so an acceleration of kx (linear) would yield a velocity of (kx^2)/2, which is a quadratic, as in C or D. Alternately, in differentionspeak, a quadratic equation for velocity as in C or D would yield a derivative that would be linear, b/c the derivative of x^2 is 2x. Again, technically, no calculus is required, but if you've seen calculus before, this may be more straightforward.
Thanks for reaching out, and hope this helps!!
This is a tough question, b/c it's a great example of how the MCAT doesn't technically test calculus, but it tests the ideas underlying calculus. The passage says that force decreases linearly with speed, which means that acceleration decreases linearly with speed. But the question is asking us to look at a graph of velocity. Keeping in mind that acceleration is the change in velocity over time, or the slope on a graph of V vs. t, we need to look for the graph that has a slope that changes linearly. In A and B, the graph is linear, but the slope is constant, so that's not what we're looking for. C and D have a slope that changes as a linear function of time.
The above explanation doesn't require calculus, but if you have taken calculus, what this boils down to is that you integrate acceleration to get velocity, so an acceleration of kx (linear) would yield a velocity of (kx^2)/2, which is a quadratic, as in C or D. Alternately, in differentionspeak, a quadratic equation for velocity as in C or D would yield a derivative that would be linear, b/c the derivative of x^2 is 2x. Again, technically, no calculus is required, but if you've seen calculus before, this may be more straightforward.
Thanks for reaching out, and hope this helps!!
Andrew D.
Content Manager, Next Step Test Prep.
Content Manager, Next Step Test Prep.