Answer
(a) $12.86~m$
(b) $21.72~m$
(c) $30.58~m$
(d) $Separation=4~m+(8.86~m/s)t$
Work Step by Step
See the picture on page 41. The origin is at the bridge and the positive direction is downward.
Now, let's find the velocity of the first rock after it has fallen $4~m$.
$v_0=0$, $a=g=9.81~m/s^2$, $\Delta x=4~m$
$v^2=v_0^2+2a\Delta x$
$v^2=0^2+2(9.81~m/s^2)(4~m)$
$v=±\sqrt {2(9.81~m/s^2)(4~m)}$. The rock is moving downward (positive direction).
$v=8.86~m/s$
Now, the second rock is dropped. Let's find the position as function of time for both rocks. The initial time is when the second rock is dropped.
- Rock 1: $x_{0_1}=4~m$, $v_{0_1}=8.86~m/s$, $a=g=9.81~m/s^2$
$x_1=x_{0_1}+v_{0_1}t+\frac{1}{2}at^2$
$x_1=4~m+(8.86~m/s)t+\frac{1}{2}(9.81~m/s^2)t^2$
$x_1=4~m+(8.86~m/s)t+(4.905~m/s^2)t^2$
- Rock 2: $x_{0_2}=0$, $v_{0_2}=0$, $a=g=9.81~m/s^2$
$x_2=x_{0_2}+v_{0_2}t+\frac{1}{2}at^2$
$x_2=0+0t+\frac{1}{2}(9.81~m/s^2)t^2$
$x_2=(4.905~m/s^2)t^2$
(a) $t=1.0~s$
$x_1=4~m+(8.86~m/s)(1.0~s)+(4.905~m/s^2)(1.0~s)^2$
$x_1=17.765~m$
$x_2=(4.905~m/s^2)(1.0~s)^2=4.905~m$
$Separation=x_1-x_2=17.765~m-4.905~m=12.86~m$
(b) $t=2.0~s$
$x_1=4~m+(8.86~m/s)(2.0~s)+(4.905~m/s^2)(2.0~s)^2$
$x_1=41.34~m$
$x_2=(4.905~m/s^2)(3.0~s)^2=19.62~m$
$Separation=x_1-x_2=41.34~m-19.62~m=21.72~m$
(c) $t=3.0~s$
$x_1=4~m+(8.86~m/s)(3.0~s)+(4.905~m/s^2)(3.0~s)^2$
$x_1=74.725~m$
$x_2=(4.905~m/s^2)(3.0~s)^2=44.145~m$
$Separation=x_1-x_2=74.725~m-44.145~m=30.58~m$
(d) $Separation=x_1-x_2=4~m+(8.86~m/s)t+(4.905~m/s^2)t^2-(4.905~m/s^2)t^2=4~m+(8.86~m/s)t$
We conclude that the separation between the rocks increases linearly with time.
