Question

Suppose s(t) is a twice-differentiable function that represents the linear motion of a particle. In which of the following situations would the particle be speeding up at t = 3?

Answer 1

"speeding-up" is the same thing as saying accelerating i.e. the velocity is increasing -- this means \(s''>0\), right?

Answer 2

so which of those examples exhibits \(s''>0\) at \(t=3\) :-) ?

Answer 3

b

Answer 4

Careful; It's speeding up when \(s^{\prime},s^{\prime\prime}>0\) or \(s^{\prime},s^{\prime\prime}<0\).

Answer 5

In other words, a particle speeds up when the velocity and acceleration are of the same sign.

Answer 6

Wait I am confused so does the < sign have to be in the same direction?

Answer 7

If you want to find when it's speeding up, the signs have to be in the same direction (either both are > or both are <).

Answer 8

Oh so that would mean that it is s'(3) < 0 and s″(3) < 0

Answer 9

because the < are in the same direction

Answer 10

Yes. At first, you may wonder how both being < would indicate speeding up. The reasoning for why this is the case is because when velocity and acceleration are both negative, you're moving in the opposite direction of (intended) motion.

Answer 11

ooooo that's tricky -- oops

Answer 12

Thanks!

Answer 13

Could you help me with something else too?

Answer 14

Just an additional remark: If the question was asking you to find where it's slowing down, you want to have either \(s^{\prime}>0,s^{\prime\prime}<0\) or \(s^{\prime}<0,s^{\prime\prime}>0\) (i.e. velocity and acceleration have opposite signs).

Sure, I can help out with something else. :-)

Answer 15

Here's an example of such a possible g(x):