Question

Can an object have a speed of zero while it has an acceleration that is not zero? Explain.

Answer 1

Zephyr, I wouldn't assume that gravitational acceleration is what this question is looking for. Gravitational acceleration is the 'apparent' acceleration you would be need to experience to achieve the same amount of force exerted by gravity. Standing on the ground, you have no change in velocity, so no real acceleration.

For this question, I would say to think about a ball that you throw up in the air. At the top of it's arc, it changes direction. At that instant, it has no velocity. But, since there is no force other than the force of gravity acting on it, it is still accelerating.

Answer 2

@zephyr141

what are you mentioned is horribly wrong.
when you stand still and do nothing, you have zero acceleration. only when you are in free fall.. do you get a 9.8 m/s^2 acceleration :P

Answer 3

Although to ad to ZippySlug's answer at the highest point, i.e. where it stops, the acceleration is zero(although, if you know anything about calculus this is theoretically impossible, since it can only approach zero).

Answer 4

@BPDlkeme234
" i.e. where it stops, the acceleration is zero"
what nonsense is that..? :-/

Answer 5

BPDlkeme, the acceleration is NOT zero at the top point. There is still a force on the object. F=ma, so if the mass is finite, and there is a force, there is acceleration. The 2nd derivative of position is constant in freefall, even at the highest point.

Further, the intermediate value theorem states that the velocity MUST be zero as it transitions between positive and negative velocity.
http://en.wikipedia.org/wiki/Intermediate_value_theorem
Where are you getting this 'if you know anything about calculus' idea? I would be interested in your credentials to make such a statement.

Answer 6

@ZippySlug .. no ..he makes no sense :P

Answer 7

ZippySlug, I dont understand, you say on the one hand yu disagree that acceleration is zero, and on the otherhand prvide proof that it is zero (intermediate value theorem).

Answer 8

In relation to calculus, delta x of dv/dt always approaches x but never reaches x; similar for a capacitor discharging, it is theoretically never discharged, although in practice it is discharged at 23 T.