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?

Here is my thought process:

Here is what I know:

When t=0 $v_i=0 ft/sec$ We also know $x_0=0 $ ft We are told $x_1-x_0=1$ ft and that the acceleration is constant. This allows us to use the formula: $x = x_0 + v_0t + \frac{1}{2}at^2$ to solve for the constant acceleration. So we get

$1 = 0ft + 0ft/s * (1s-0s) + \frac{1}{2}a(1s-0s)^2$. Doing the math gives

$a = 2(ft/s)/s$ Now we know the constant acceleration. We can use this value to find the velocity at t=1 second by applying the formula:


$v=0+2(ft/s)/s * (1-0) = 2 ft/s$ Therefore the ball is rolling with a velocity 2 ft/s at time t=1.

Finally, I can use :

$x = x_0 + v_0t + \frac{1}{2}at^2$ again to solve for the distance the ball traveled after 2 seconds:

$x_2=x_1 + v_1t + \frac{1}{2}at^2 = 1ft + (2 ft/s)*(2s-1s) + \frac{1}{2}*(2ft/s/s)(2-1)^2= 1ft + 2ft + 1ft = 4ft$ Therefore the ball has traveled 4 feet in the first 2 seconds. Since the question only asks for the distance traveled between t=1 and t=2 I need to subtract out the distance traveled in the first second (1 ft). 4-1 = 3ft.

Therefore the ball traveled 3 feet from t=1 to t=2.

Is this correct?


closed as off-topic by John Rennie, HDE 226868, ACuriousMind, user10851, David Z Sep 3 '15 at 5:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, HDE 226868, ACuriousMind, Community, David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Ernie - can you should me how you are getting acceleration to be 1ft/s/s ? Because I am getting 2 ft/s/s. $\endgroup$ – user1068636 Sep 2 '15 at 2:39
  • $\begingroup$ Acceleration during the first second: a = (V final - V initial) / t = 1 second per second. As acceleration is constant, the velocity during the 2nd second is 1 + 1 = 2 ft./sec. Therefore it moved 2 feet from t=1 to t=2. Unless I'm missing something really obvious. $\endgroup$ – Ernie Sep 2 '15 at 2:52
  • $\begingroup$ Ernie - using the formula $x = x_0 + v_0t + \frac{1}{2}at^2$ above I got a=2. What did I do wrong? $\endgroup$ – user1068636 Sep 2 '15 at 3:16
  • $\begingroup$ a does equal (V final - V initial)/t but Ernie miss-calculated somehow because V final is 2ft/s and V initial is 0ft/s therefor a = 2ft/s^2. $\endgroup$ – Alex Sep 2 '15 at 3:52
  • 1
    $\begingroup$ Yes, you and Alex are right. Average velocity during the first second is 1 ft/sec, so final velocity at t=1 must be 2 ft/sec as the ball has constant acceleration. So acceleration is 2 ft/sec^2. $\endgroup$ – Ernie Sep 2 '15 at 5:42

Take the kinematic relationship $$ x_{end} - x_{start} = \frac{ v_{end}^2 - v_{start}^2 }{2 a }$$ which applies to constant acceleration and use it twice.

First on the first interval to calculation the acceleration $a$ and then again on the second interval to find the distance traveled.


Yes, by my calculations with a = 2ft./s^2
Just use x(t) = 1/2at^2
x(2) = 4 feet
x(1) = 1 foot
So, x(2) - x(1) = 3 feet


Not the answer you're looking for? Browse other questions tagged or ask your own question.