Simple kinematics question i think i am right just need to confirm

[ihatecats2014]

Homework Statement

At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1 foot between t=0 seconds and t = 1 second.
How far does it move between t = 1 second and t = 2 seconds

Homework Equations

delta x=v_it+1/2at²? since it is constant speed wouldn't acceleration be 0?

The Attempt at a Solution

it seems like it would be 1 foot right?
= 1(1)+1/2(0)(1)²
delta x=1foot?

 

[rock.freak667]

No the acceleration would not be zero. If it were zero, then the ball would not move.

Homework Statement

At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1 foot between t=0 seconds and t = 1 second.

Using this information and the same relevant equations, what is the acceleration? (Remember, initially at rest means no initial velocity)

 

[ihatecats2014]

No the acceleration would not be zero. If it were zero, then the ball would not move.
Using this information and the same relevant equations, what is the acceleration? (Remember, initially at rest means no initial velocity)

would the acceleration be 1? because it went 1 foot in the elapsed time of 1 sec?

then
=0(1)+1/2(1)(1)^2
delta x equals 1/2 foot
this is wrong. i need a hint please

or if the time is 2sec the distance would then be 2 foot. but i think that is still wrong

 

[rock.freak667]

would the acceleration be 1? because it went 1 foot in the elapsed time of 1 sec?

Your equation is x=ut+0.5at² (u= initial velocity)

Using this equation a≠1 ft/s².

 

[rwisz]

Your attempt at a solution is correct. Since the ball is moving with constant acceleration, the equation for displacement is simply delta x = vit + 1/2at^2. However, since the acceleration is 0, the second term in the equation becomes 0, leaving us with delta x = vit. Since the initial velocity is also 0, the displacement between t = 1 second and t = 2 seconds is also 0. Therefore, the ball does not move any further during this time interval.