One year it just feels like less because you’re going so fast
Displacement = final - initial
The formula most closely resembling that is delta = xf - xi
Answer:
The velocity must change but not speed.
Explanation:
- Velocity is defined as the displacement by time. Whereas speed is expressed as the distance between two successive positions of the body to the time interval it took to travel.
<em>Velocity, V = D / t m/s</em>
<em> Speed, s = d /t m/s </em>
- Velocity is a vector quantity that has a magnitude and direction.
- The speed is a scalar quantity having only the magnitude.
- At any instant of time, the magnitude of the velocity is always equal to the magnitude of the speed. The magnitude of velocity, |<em>v </em>| = magnitude of speed, |<em>v </em>|. The magnitude is always positive
- The acceleration of a body is defined as the rate of change of velocity to time.
<em> a = (v - u) / t m/s²</em>
- If a body is accelerating, It varies its velocity with respect to time.
- In case of uniform circular motion, the speed remains constant, but the velocity changes continuously.
So, in the case of circular motion if an object accelerates, velocity must change but speed remains constant.
The block's speed at the point where x=0.25A is v = 31.95 cm/s.
<h3>What is Spring constant?</h3>
The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.
question is incomplete, this is the remaining statement
What is the amplitude of the subsequent oscillations? And What is the block's speed at the point where x=0.25A?
x = Asin(wt)
v = Aw coswt
at t = 0
w = sqrt(k/m)
v = Aw
A = v/w
A = 7.17 cm
part b )
E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
mv^2 + k(1/4A)^2 = 1/2kA^2
mv^2 + kA^2/16 = kA^2
mv^2 = kA^2 - kA^2/16
mv^2 = 15kA^2/16
v^2 = 15/16 * (k/m) * A^2
v^2 = 15/16 *w^2A^2
v = sqrt(15/16) * wA
v = 31.95 cm/s
to learn more about spring constant go to -
brainly.com/question/23885190
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