How Does a Rolling Ball's Velocity Change with Constant Acceleration?

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The discussion focuses on calculating the velocity, average velocity, and displacement of a rolling ball with an initial velocity of -1.5 m/s and a constant acceleration of -0.23 m/s² over a time interval of 2.2 seconds. The final velocity after 2.2 seconds is determined to be -2.006 m/s using the formula V = initial velocity + acceleration × time. The average velocity during this interval is calculated as the average of the initial and final velocities, while displacement can be found using the kinematic equation s_f = s_i + v_i t + 1/2 a t².

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Question: A rolling ball has an initial velocity of -1.5 m/s.
a. If the ball has a constant acceleration of -0.23 m/s^2, what is its velocity after 2.2s?
b. What was its average velocity during that time interval?
c. How far did it travel in this time interval?

So far i got:
a. V= initial velocity + acceleration(time)
(-1.5m/s) + (-0.23m/s^2)(2.2s)=-2.006m/s

b. i know the formula for average velocity V= (d2- d1) / (t2 -t1). My problem is I'm not sure what to plug in for displacement. for time i plugged in 2.2s -0s this is where i am stuck.

c. I am not sure of the formula to solve this part.
 
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Calculate displacement using simple kinematic relations.

Use v2 you calculated above.

You need not pay much attention to the negative sign given in velocity.it only indicated that the velocity is in the backward direction as from the general Cartesian conventions.
don't plug in the negative sign to get the value of velocity.

For next part you can use simple kinematic equations.

v[initial]= 1.5m/s
accln= -.23 m/s^2
 
Last edited:
Sry, but this still leaves me with the same problem I already figured out question a. I know the formula for question b, but my problem is I don't know what to put for displacement. How would i figure out d1 and d2 in the formula. For question c I am not sure how to solve this question so how would i begin.
 
Jess048 said:
Question: A rolling ball has an initial velocity of -1.5 m/s.
a. If the ball has a constant acceleration of -0.23 m/s^2, what is its velocity after 2.2s?
b. What was its average velocity during that time interval?
c. How far did it travel in this time interval?

So far i got:
a. V= initial velocity + acceleration(time)
(-1.5m/s) + (-0.23m/s^2)(2.2s)=-2.006m/s

b. i know the formula for average velocity V= (d2- d1) / (t2 -t1). My problem is I'm not sure what to plug in for displacement. for time i plugged in 2.2s -0s this is where i am stuck.

c. I am not sure of the formula to solve this part.

If you figure out c, you'll have calculated d2-d1.

I'm surprised you'd know the equation you used in a without knowing the equation for determining your position. The equation you used in a is just the derivative of your equation for position: s_f = s_i + v_i t + 1/2 a t^2
 
Jess048 said:
i know the formula for average velocity V= (d2- d1) / (t2 -t1). My problem is I'm not sure what to plug in for displacement.

Hi Jess048! :smile:

You don't need displacement for this part.

For constant acceleration, the formula for average velocity is also (Vi + Vf)/2 … in other words, it's the average of the initial and final velocity.

The formulas do give the same result:

(df - di)/(tf - ti) = (atf^2/2 - ati^2/2)/(tf - ti) = (atf + ati)/2 = (Vf + Vi)/2. :smile:
 

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