Physics help needed?

Question

I am trying to learn how to solve the following problem. Thanks for your help:

If a rolling ball has an initial velocity of -1.5 m/s and a constant acceleration of -0.23m/s^2 what is the average velocity after 2.2s?

Thanks for your help. Serious answers only, I am trying to learn how to do this from a book that is not real clear.

How far did it travel druing this time?

Answer 1

-0.23m/s^2 times 2.2 s = -0.506m/s + (-1.5) = -2.006 that is your final velocity
average velocity equals final velociy plus intial velocity divided by 2. Which means
average velocity = [-2.006 + (-1.5)] / 2
= [-2.006 - 1.5] / 2
= -3.506/2
= -1.753 m/s

To See how far it has travled use the formula
v2^2 = v1^2 + 2a*(s2 - s1)
v2 = final velocity
v1 = intial velocity
a = acceleration
s2 = final position
s1 = intial position

First let's list the knowns
Known
v2 = -2.006 m/s
v1 = -1.5 m/s
a = -0.23 m/s^2
s1 = 0

Unknown
s2

Now we have to solve v2^2 = v1^2 + 2a*(s2 - s1) to get s2 by it self to do this
v2^2 = v1^2 + 2a*(s2 - s1) first subtract v1^2 from both sides and we get
(v2^2 - v1^2) = 2a*(s2 - s1) now we have to divide both sides by 2a which gives
[(v2^2 - v1^2) / 2a] = s2 - s1 now add s1 to both sides which gives
s2 = [(v2^2 - v1^2) / 2a] + s1 now to solve let's plug in our knowns we get
s2 = [((-2.006)^2 - (-1.5)^2)) / 2(-.23)] + 0
= ((4.024036 - 2.25) / -.46)
= -3.8566 m. But distance can not be negative so it is 3.8566 m away. ( The negative suggests it is to the left since on a number line numbers decrease towards the left)

If you need further clarification or more help email me or im me. Good Luck.

Answer 2

Plugging into these equations will give you the answer.

Velocity (final) = Velocity (initial)+Acceleration x Time
OR
v(f) = v(i)+at

Basically the equation just takes the Initial velocity and adds the acceleration over time (a*t) and that gives you the resultant velocity. Now, with that resultant velocity and the initial, we can find the average velocity.

Velocity (Average) = .5 * (velocity(initial) + velocity(final))

As for how far it traveled, due to the way the question is phrased, you have to use the Displacement Formula.

Displacement (or change in distance) =
Velocity (average) * time

(Delta [the triangle])X = V(average)*t

In this equation, the Delta-X, or change in X (which is the distance), is simply obtained by multiplying the average velocity of the object with the time. The math leaves us with the displacement from the initial spot, or the distance it traveled.

This is a great site to use for these types of problems. It has a list at the bottom of good equations.

http://www.ajdesigner.com/constantacceleration/cavelocity.php

*ANSWER BELOW*








V(final) = -2.006m/s
V(average) = -1.753m/s
Delta-X = -3.8566 meters (3.86 meters backwards)

Answer 3

acceleration equals (final velocity - initial velocity) divided by time.

a= (v-u)/t

In your question,
u= -1.5m/s
v= ?
t= 2.2s
a= -0.23m/s^2

So, using the formula we get...

a= (v-u)/t
-0.23m/s^2 = (v- -1.5)/2.2
-0.23m/s^2= (v +1.5)/2.2

(Remember when subtracting a nrgative you get a positive)

We can then rearrange the above equation...

2.2 x -0.23 = v + 1.5
-0.506 = v + 1.5
-0.506 -1.5 = v
v= -2.006

v is the final velocity which has been calculated from the above equation to be 2.006 m/s.

Average velocity is given by the formula : (u+v)/2

Using this formula we get,

Avg. V = (-1.5 + -2.006)/2
= -1.753 m/s

The average velocity is -1.753m/s


To find the distance travelled during this time you can use the following formula :

s = (u x t) + 0.5 x a x (t^2)

Where 's' is the displacement (distance travelled in a set direction), a is accerleration, u is initial velocity and t is time.

Using the formula we get,,,

s = (-1.5 x 2.2) + 0.5 x -0.23 x (2.2^2)
= -3.3 + (-0.115 x 4.84)
= -0.5566m

Thus, the distance travelled is 0.5566m

I hope this helps!!

Answer 4

The defining equation is
v = v0 + at (final velocity is initial velocity plus acceleration times time) so
-1.5 + (-.23*2.2) = -2.006 m/s

Average velocity is the average of the initial and final velocities, so
(-2.006+ (-1.5))/2 = -1.753 m/s

HTH

Doug