physics questions?

Question

1) Under what conditions would acceleration be different from the "rate of change of speed"?
2)Under what conditions could an object continue to travel at a constant speed and yet be accelerated?
3)A ball is thrown straight up and is observered to come straight down. How does the acceleration of the falling ball compare to its acceleration on the way up? What is the acceleration at the very peak of the motion when its velocity is zero?
4)What would happen to the height, acceleration, return velocity, and time of flight of a vertically thrown ball if its initial upward velocity was doubled?

Answer 1

1. Acceleration is a vector, so in case of angular velocity (where the direction always changes) acceleration will be changing all the time as well (the scalar value will remain same though)

2. When there is no external force inhibiting movement (e.i gravity, air resistance etc) an object will continue to tarvel at a constant speed. Acceleration is the rate of change of velocity with respect to time. Now in this case velocity is fixed, so to get acceleration we have to change time (hypothetically), so in case of Time travel (as in science fiction) if an object is moving at a constatnt velocity it will have an acceleration (due to changing dimention of time)

3. When the ball is going up, it's velocity is gradually reducing (due ot the Earth's gravity). So the rate of change of velocity is negative so it will have negative acceleration which is commonly called deceleration. When the ball is falling the change in velocity is positive (as the veocity will gradually increase due to earth gravity) so it is called actually acceleration.

The acceleration of the ball going up and ball falling from down are same only the sign is different (going up is deceleration, going down is acceleration)

At the peak velocity is zero so rate of change of velocity is also zero. Thus at the peak acceleration is zero.

4. okay now you follow this formula

H (height)
g (acceleration due to gravity)
V (final velocity) it will be zero for the ball thrown up when it reaches the height H
U (Initial velocity)

V*V=U*U-2gH

or 0 = U*U - 2gH
or H = (U*U)/2g

Now if the initial speed is doubled the say the ball reaches a new height H1

So H1 = (2U*2U)/2g = 4 ((U*U)/2g) = 4H

So if you throw the ball with double speed it will reach 4 times the initial height.

Acceleration will remain same (g). It doesn't depend on the velocity you throw, rather dependednt on the gravity

Now using the same formula you can calculate the return velocity (I mean the velocity at which it will hit the ground)

Now the formula will be

H (height)
g (acceleration due to gravity)
V (final velocity) it will be the velocity at which it will hit the ground
U (Initial velocity) which is zero

V*V=U*U-2gH

so V*V = 2gH
so V= sqrt (2gH)

Now if the Height becomes 4 times (due to doubling of the velocity the ball was thrown up), say the new velocity becomes V1

so V1 = sqrt (2gH1) = sqrt (2.g.4H) = 2. sqrt (2gH) = 2 V

So the velocity at which it will hit the ground will be doubled

Now consider another formula to calculate the time of flight

V (Final velocity) while throwing the ball it will be zero at height H
U (Initial velocity)
H (Height the ball will reach)
g (acceleration due to gravity)
t (the time it will take to reach to H)

V = U - gt
so 0 = U - gt
so t = U/g

While coming down

V = U - gt (here V is the velocity at which the ball will hit the ground and U is the initial velocity which is zero at height H)

V = gt
so t = V/g

(Note: the ball will hit the ground at the same velocity it was thrown up)

So total time of flight is 2V/g. Now if the initial velocity is doubled, the time of flight will become 2(2V)/g = 2t.

So flight time will also be doubled

Summary: if the velocity at which the ball is thrown up is doubled it will
a. reach 4 times height
b. double the return velocity
c. double the flight time

Hope this come to your help.

Answer 2

1)U know that acceleration is defined as rate of change in VELOCITY.So, consider UNIFORM circular motion. The velocity is always changing since the direction is always changing which means there is some acceleration-variable or constant(although it is not of our concern).But speed is constant and if we defined acceleration as rate of change of speed,it would be 0 but there is some acceleration.Hence the contradiction
2)In the uniform circular motion.The velocity is always changing since the direction is always changing which means there is some acceleration-variable or constant(although it is not of our concern)
3)the acceleration throuout the journey is constant and equal to g
provided the herght the ball is thrown is VERY small compared to the radius of earth.Otherwise,it would keep changing and te acceleration woule vary with height as
a is proportional the inverse square of the distance from the ball to the cetre of earth.
whatever it may be the acceleration has a downward direction.
4)if the height is very small,and velocity is doubled and air drag is neglected,
*height would increase by 4 times H=v^2/2g
*acceleration would still be g (provider height is very small)
*retuun velucity would be the same velocity with which u threw up the ball(in the particular case)
*time of flight would be double since T=2v/g

Answer 3

1. object in circular motion.

2. object in circular motion

3. acceleration of falling ball is +g(9.8m/s^2). i.e towards center of earth in the direction of motion
acceleration of ball while thrown up is -g i.e.towards center of earth in a direction opposite to that of motion
at all points during motion. accel is constant g. At the peak only the negative sign becomes positive.

4. if velocity is doubled height increases 4 times, acceleration is answered in part 3, return velocity is numerically equal to throwing up,(direction is different). Try calculating time yourself you might want to use any of these..
v = u - at
v^2 = u^2 - 2gs
s = ut - 1/2 gt^2

Answer 4

1.) if it's angular acceleration, then you have a change in direction, which is a part of your velocity, but a different part than speed.

2.) if it's rotating in a circle

3.) acceleration is always constant, whatever the gravitational acceleration is on whatever body you're on

4.) all of those would be doubled except acceleration, which is always constant

Answer 5

1 and 2)if an object changed direction
3)It has negative acceleration going up and positive acceleration going down. At the peak, its acceleration is 0m/s/s.
4)not sure, sorry

Answer 6

1) dunno
2) dunno
3) a)may be higher since the retun velocity is restricted by air friction (maximum velocity)
b) 0
4) return velocity depends on how high it goes, if it's 2 ft, the difference is negligible. If it's 1,000 feet, the difference is much great since return maximum velocity will be the same either way, but upwards velocity will be much different.

Answer 7

i'm at the moment a pupil in college taking the direction Physics 4 with Calculus. i got here upon intense college Physics to be somewhat person-friendly. it really is elementary for most intense schoolers. AP Physics is likewise person-friendly. in case you want technology and math, then you definately might want to do not have any difficulty taking preAP Physics. it really is way a lot less complicated than college Physics with Calculus.

Answer 8

Those are physics questions? Physics is pretty simple isn't it?

Answer 9

are you doing homework? sorry i can't help school was out one week ago. :( good luck

3. they are the same.