As an object goes round in a circular path, then its velocity will along the tangent at that instant. But centripetal acceleration is normal to that tangent and so along the radius of curvature. As acceleration is perpendicular to the velocity, the direction aspect is ever changing and so the object goes round the circular path.
acceleration is the rate of change of velocity as a function of time
Acceleration.If the velocity is "v" and the distance (here, I suppose you meant displacement) is "d", then we get:[v^2/d]=[(m/s)^2/m]=m^2*s^2/m=m/s^2And that is the S.I. unit of acceleration.If you learn Physics, pay attention to the written below this sentence.There are some cases that allow you to know what is the coefficient also (coefficient of 2x, for example, is 2). If an object moves kinematically, then you know it if the acceleration is known to be constant:Constant speed=0 acceleration (then, 0 m/s^2).Constant acceleration (different than 0): If Vf is the final speed of the moving body, Vi is the initial speed and "a" is its acceleration, then we know that: d=(Vf2-Vi2)/2a. Suppose that initial speed is zero, so we get: v^2/d=2a, so in kinematic movements where the acceleration is not changing, the coefficient of your slope is 2. In both cases, the graph representing that ratio is linear.
A fictitious force caused by rotation - it feels as if a force pushes you towards the outside. The magnitude of the ficticious acceleration is equal to the real centripetal acceleration: a = v2/r. The corresponding force can be obtained from Newton's Second Law.
To be very technical regarding the language of the question, No.The question specifies "constant velocity" ... meaning constant speed AND direction.If either of these is changing, then acceleration is present.But if the question meant to say "constant SPEED", then the answer is Yes. An object may be traveling at constant speed, but if its path is not a straight line, then there is acceleration present.
it is just like gravitational effect on a rocket it is the max vel required to overcome the circular motion tangential to the centripetal force
acceleration is the rate of change of velocity as a function of time
the rate of change in the velocity of a body
The answer becomes clear with the aid of some simple mathematics. Acceleration is defined as (change in velocity)/(time for the change) . Clearly, then, acceleration is infinite whenever velocity changes in "no time".
Acceleration.If the velocity is "v" and the distance (here, I suppose you meant displacement) is "d", then we get:[v^2/d]=[(m/s)^2/m]=m^2*s^2/m=m/s^2And that is the S.I. unit of acceleration.If you learn Physics, pay attention to the written below this sentence.There are some cases that allow you to know what is the coefficient also (coefficient of 2x, for example, is 2). If an object moves kinematically, then you know it if the acceleration is known to be constant:Constant speed=0 acceleration (then, 0 m/s^2).Constant acceleration (different than 0): If Vf is the final speed of the moving body, Vi is the initial speed and "a" is its acceleration, then we know that: d=(Vf2-Vi2)/2a. Suppose that initial speed is zero, so we get: v^2/d=2a, so in kinematic movements where the acceleration is not changing, the coefficient of your slope is 2. In both cases, the graph representing that ratio is linear.
If the velocity is uniform, then the final velocity and the initial velocity are the same. Perhaps you meant to say uniform acceleration. In any event, the question needs to be stated more precisely.
A fictitious force caused by rotation - it feels as if a force pushes you towards the outside. The magnitude of the ficticious acceleration is equal to the real centripetal acceleration: a = v2/r. The corresponding force can be obtained from Newton's Second Law.
To be very technical regarding the language of the question, No.The question specifies "constant velocity" ... meaning constant speed AND direction.If either of these is changing, then acceleration is present.But if the question meant to say "constant SPEED", then the answer is Yes. An object may be traveling at constant speed, but if its path is not a straight line, then there is acceleration present.
it is just like gravitational effect on a rocket it is the max vel required to overcome the circular motion tangential to the centripetal force
9.8 m/s squared is the acceleration due to gravity on earth. So if you drop any object from a certain point it will accelerate at 9.8 m/s squared down.
Dropping a stone from a tall building is an example of acceleration due to gravity. The stone's speed will increase as it falls until it reaches terminal velocity.
Perhaps you meant speed or velocity, because the vertical acceleration is constant throughout the bomb's decent, ignoring the effects of air resistance. The acceleration due to gravity is 9.8 m/s2 for all values of t.
Velocity is a speed in a very specific direction.