Physics Exercises/Kinematics in One Dimension/Solutions

Without Calculus

1. 3.1 s.

   Use the kinematics equation $y=y_0+v_{0}t+(1/2)a{t}^2$. Let the rock's initial height $y_0=0$. Then equation simplifies to $y=t(v_0+(1/2)at)$. So, two solutions for t are: t = 0 s and $t=-2v_0/a$. Plugging in $v_0=15\mathrm{m}/\mathrm{s}$ and $a=-9.8\mathrm{m}/{\mathrm{s}}^{\mathrm{2}}$ gives t = 3.1 s.

2. Let L be the length of the escalator, v₁ be the speed of the escalator, and v₂ be the speed the man walks on solid ground. It takes t₁ for a man standing on the escalator to travel its length, so $L=v_1{t}_1$. If the man walks against the escalator, his net speed is $v_2-v_1$; it takes him t₂ to travel the length of the escalator, so $L=(v_2-v_1)t_2$. Since $L=v_1{t}_1=(v_2-v_1)t_2$, we find that $v_1(t_1+t_2)=t_2{v}_2$ or $v_2=\{(t_1+t_2)/t_2\}{v}_1$. Now, when the man walks down the escalator, his net speed is $v_1+v_2$. If it takes him t seconds this time, then $L=(v_1+v_2)t=\{1+(t_1+t_2)/t_2\}{v}_{1}t$. Rewriting $v_1$ in terms of given quantities, $v_1=L/t_1$, and solving for t, we get:
   $t=(t_1{t}_2)/(t_1+2t_2)=3.75\mathrm{s}$.

   Note that by carrying out the calculation in symbols as far as we can, we actually discover that the final answer is independent of L. This is one of many benefits of not substituting in numbers until the very end. (As with any rule, this rule has exceptions, but start by following the rule—you will notice exceptions by yourself when you become sufficiently skilled.)

With Calculus

1. The basic equations are these: $a=\frac{dv}{dt}$, $v=\frac{dx}{dt}$. Since our final goal is x as a function of t, we should solve for v(t) first by integrating the first expression:

   $v(t)=\underset{0}{\overset{t}{\int }}dv=\underset{0}{\overset{t}{\int }}ad{t}^{\prime }=at+v_0$

   And integrating this a second time yields the final expression for x(t)

   $x(t)=\underset{0}{\overset{t}{\int }}dx=\underset{0}{\overset{t}{\int }}vd{t}^{\prime }=\underset{0}{\overset{t}{\int }}(at+v_0)d{t}^{\prime }=(1/2)a{t}^2+v_{0}t+x_0$.

2. e cm.

   The positions as a function of time can be written down as the infinite series:

   $x=x_0+v_{0}t+(1/2)a_0{t}^2+(1/3\cdot 2)j_0{t}^3+\cdots$.
   Once you let $x_0=v_0=a_0=j_0=\cdots =1$, you should recognize the resulting series as the Taylor series of $e^t$ (times 1 cm):
   $e^t=1+t+(1/2)t^2+(1/3!)t^3+\cdots$.

   If you plug in $t=1$, you get $x=e^1=e\mathrm{c}\mathrm{m}$.

   Observant reader should notice that we suddenly dropped units in the middle. To be thorough, we would factor out "1 s" from each written expression of t and absorb it into the coefficient (that way, t becomes unitless and all coefficients have unit of cm).