When an object is accelerating, does speed, velocity, or force change?

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When an object accelerates, its velocity can change even if its speed remains constant, particularly in circular motion. This is because velocity is a vector quantity that includes both magnitude and direction, while speed is only a measure of magnitude. Examples such as a tetherball on a string or a car navigating a curve illustrate that constant speed can coincide with changing velocity. Acceleration encompasses speeding up, slowing down, or changing direction, which is crucial for understanding motion dynamics. Therefore, the distinction between speed and velocity is essential in physics.
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qn3: why is statement 2 the only answer? isn't speed one of the factors in velocity too? then shouldn't statement 1 be true too?
 
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Velocity can change without speed changing. This occurs in circular motion, where the velocity vector changes continuously while speed is held constant. Examples include a tetherball on a string, a car going around a curve at a constant speed, and an object in a circular orbit.
 
You have to remember that in physics and maths, acceleration means:
1. speeding up or
2. slowing down or
3. changing direction
(or a mixture of 1&3 or 2&3).
That's because velocity is a vector (has magnitude and direction) and acceleration is how fast velocity changes.
(This is different to the non-scientific use of the word 'acceleration' which simply means speeding up.)

So, as @Drakkith notes, for something going in a circle at a steady speed, the velocity is changing (as the direction of velocity is constantly changing). This is classified as 'acceleration'.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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