Velocity Changes: Understanding Direction and Acceleration in Changing Objects

  • Thread starter Thread starter UrbanXrisis
  • Start date Start date
  • Tags Tags
    Velocity
AI Thread Summary
When an object changes velocity from -4 m/s to -1 m/s, it is becoming less negative, thus closer to zero, which means it is less positive. In the second scenario, where the object shifts from 2 m/s to -2 m/s, the velocity is also becoming less positive as it transitions from a positive to a negative value. Both scenarios involve acceleration; the first represents negative acceleration due to slowing down, while the second involves a change in direction, indicating centripetal acceleration. Understanding these changes in velocity is crucial, as they always signify acceleration regardless of whether the velocity is increasing or decreasing. Changes in velocity, whether in speed or direction, consistently indicate the presence of acceleration.
UrbanXrisis
Messages
1,192
Reaction score
1
What if an object is changing from -m4/s to -1m/s. Is the velocity becoming more positive or less positive? Is there acceleration in this?

What if an object is turning around so that it undergoes a change in velocity from 2m/s to -2m/s? Is the velocity becoming more positive or less positive? Is there acceleration while the object is turning?
 
Physics news on Phys.org
UrbanXrisis said:
What if an object is changing from -m4/s to -1m/s. Is the velocity becoming more positive or less positive?
I assume you mean -4 m/s to - 1 m/s. Well, you figure it out. Draw a number line from -10 to +10 and mark off where -4 and -1 is. When you move from one to the other, which direction do you move, positive or negative.
Is there acceleration in this?
If the velocity changes, there's acceleration.

What if an object is turning around so that it undergoes a change in velocity from 2m/s to -2m/s? Is the velocity becoming more positive or less positive? Is there acceleration while the object is turning?
Figure this one out the same way.
 


In both of these scenarios, the velocity is becoming less positive. In the first scenario, the object is slowing down from -4m/s to -1m/s, so its velocity is becoming less negative and therefore closer to 0, making it less positive. In the second scenario, the object is changing direction and its velocity is going from a positive 2m/s to a negative 2m/s, so again, its velocity is becoming less positive.

There is acceleration in both scenarios. In the first scenario, the object is slowing down, which means it is experiencing a negative acceleration. In the second scenario, the object is changing direction, which also requires a change in velocity, and therefore acceleration. This acceleration is known as centripetal acceleration, as it is directed towards the center of the circular motion the object is undergoing while turning.

Overall, it is important to understand that changes in velocity, both in terms of speed and direction, indicate acceleration. Whether the velocity is becoming more positive or less positive depends on the specific scenario and the direction of the change. But regardless, there is always acceleration involved in these types of velocity changes.
 
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