Is the Velocity of an Upward-Moving, Slowing Rock Still Positive?

  • Thread starter Thread starter oreospls
  • Start date Start date
  • Tags Tags
    Velocity
AI Thread Summary
When a rock is thrown upward, its position is positive as it moves above the ground, which is defined as the origin. The velocity remains positive because the rock is still moving upward, despite slowing down. The discussion highlights the importance of defining the positive direction in physics problems, as it can lead to confusion. Participants agree that while the rock's velocity is positive, the problem should clarify the positive direction. Overall, understanding the distinction between positive velocity and the rate of change is crucial in this context.
oreospls
Messages
3
Reaction score
0
You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is___ and the velocity of the rock is___.

I said that they are both positive because the rock is going up, and even though it is slowing down, it is still going up so the velocity is positive. Is my reasoning correct?
 
Physics news on Phys.org
I think it's looking for either an "increasing" or "decreasing" (not a "positive" or "negative")
 
Nathanael said:
I think it's looking for either an "increasing" or "decreasing" (not a "positive" or "negative")

The options show positive and negative.

I think it does make sense though, because the ball is in a positive direction relative to the ground, and the velocity is upwards, but I'm not completely sure.
 
They said the ground was the origin, but they didn't say which direction (up or down) was positive.

Your answers and reasoning are correct, but the problem really should have stated which direction is the positive direction.
 
Ah I see. Thanks!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top