Physics Exercises/Kinematics in One Dimension/Solutions

Without Calculus
The speed of Cinderella,$$ v_c $$ is 12mph The time used by Cinderella,$$ t_c $$ is 0.25h The speed of Prince,$$ v_p $$ is unknown The time used by Prince,$$ t_p $$ is 0.16666h or $$1/6h$$ Since they have to meet, their distance traveled will be equal, then $$ v_c t_c = v_p t_p$$ $$ (12)(0.25) = v_p(1/6)$$ $$ v_p = 18 $$
 * 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 $$y = t(v_0 + (1/2) a t)$$. So, two solutions for t are: t = 0 s and $$t = - 2 v_0/a $$. Plugging in $$v_0 = 15~\mathrm{m/s}$$ and $$a = - 9.8~\mathrm{m/s^2}$$ 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: $$t = (t_1 t_2)/(t_1 + 2 t_2) = 3.75~\mathrm{s}$$. 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&mdash;you will notice exceptions by yourself when you become sufficiently skilled.)
 * 3) 14.4 s and 200 m  Let xb be the position of the bike and xc be the position of the car. We know from our kinematic equations that $$ x_f = x_i + v_i t + (1/2) a t^2 $$.  Because we are assuming that both the bike and the car are moving at a constant velocity (25 km/hr and 50 km/hr respectively) we can say that the acceleration (a) in this equation is zero (in both cases).  Thus we can write two equations of motion :  The equation for the bike is $$ x_b = x_{bi} + v_b t $$ and the equation for the car is $$ x_c = x_{ci} + v_c t $$.  Because we are interested in the point at which the two meet - we can safely say that we are looking for the point at which $$ x_b = x_c $$.  Solving this expression gives us $$ x_{bi} + v_b t = x_{ci} + v_c t $$ and thus the time which it will take for the two objects to reach the same location can be found to be $$ t = ( x_{bi} - x_{ci} ) / ( v_c - v_b ) $$.  Before we can plug in any values we need to make sure that they share the same units, so 25 km/hr becomes 6.944 m/s and 50 km/hr becomes 13.888 m/s.  Plugging in these values gives us $$ t = (100 m - 0 m) / (13.888 m/s - 6.944 m/s) = 14.4 s$$  Finally, in order to find the location at which they meet, simply plug this value for time into either of the two equations of motion, and we get $$ x = 0 + ( 13.888 m/s ) (14.4 s) = 200 m $$
 * 4) 18 mph

With Calculus

 * 1) The basic equations are these: $$a = \frac{d v}{d t}$$, $$v = \frac{d x}{d t}$$. Since our final goal is x as a function of t, we should solve for v(t) first by integrating the first expression: $$v(t) = \int_0^t dv = \int_0^t a~dt' = a t + v_0$$  And integrating this a second time yields the final expression for x(t) $$x(t) = \int_0^t dx = \int_0^t v~dt' = \int_0^t  (a t + v_0)~dt' = (1/2)a t^2 + v_0 t + x_0$$.
 * 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 $$x = e^1 = e~\mathrm{cm}$$.  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).