0
$\begingroup$

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:

$v=v_0+at$

$v=0+2(ft/s)/s * (1-0) = 2 ft/s$ 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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

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