Q#61
An airplane travels 2100 km at a speed of 720 km/h and then encounters a tailwind that boosts its speed to 990 km/h for the next 2800 km. What was the total time for the trip? What was the average speed of the plane for this trip? [Hint: Does Eq. 2–11d apply?]
Answer:
The average speed for each segment of the trip is given by
$\overline{v}=\frac{d}{\Delta t}$, so
$\Delta t = \frac{d}{\overline{v}}$ for each segment.
For the first segment,
$\Delta t_1 = \frac{d_1}{\overline{v}_1}$
$\Delta t = \frac{2100 \ km}{720 \ km/h}=2.917 \ h$
For the second segment
$\Delta t_2 = \frac{d_2}{\overline{v}_2}$
$\Delta t = \frac{2800 \ km}{990 \ km/h}=2.828 \ h$
Thus the total time is
$\Delta t_{tot}= \Delta t_1+\Delta t_2=2.917 \ h+2.828 \ h = 5.745 \ h$
The average speed of the plane for the entire trip is
$\overline{v}=\frac{d_{tot}}{\Delta t_{tot}}$
$\overline{v}=\frac{2100 \ km + 2800 \ km}{5.745 \ h}=852.9 \ km/h$
Q#62
A stone is dropped from the roof of a high building. A second stone is dropped 1.30 s later. How far apart are the stones when the second one has reached a speed of 12.0 m/s.
Answer:
Choose downward to be the positive direction and $y_0$ = 0 to be at the roof from which the stones are dropped. The first stone has an initial velocity of 0 υ = 0 and an acceleration of a = g. Eqs.
$v_y=v_{0y}+a_yt$ and $y-y_0=v_{0y}t+\frac{1}{2}a_yt^2$ (with x replaced by y) give the velocity and location, respectively, of the first stone as a function of time.
$v_1=gt_1$ and $y_1=\frac{1}{2}gt_1^2$
The second stone has the same initial conditions, but its elapsed time is $t_2=t_1-1.30 \ s$, so it has velocity and location equations as follows:
$v_2=g(t_1-1.30 \ s)$ and $y_1=\frac{1}{2}g(t_1-1.30 \ s)^2$
The second stone reaches a speed of $v_2$ = 12.0 m/s at a time given by
$t_2=\frac{v_2}{g}=t_1-1.30 \ s$
$t_1=1.30 \ s + \frac{v_2}{g}$
$t_1=1.30 \ s + \frac{12.0 \ s}{9.80 \ m/s^2}=2.524$
The location of the first stone at that time is
$y_1=\frac{1}{2}gt_1^2=\frac{1}{2}(9.80 \ m/s^2)(2.524 \ s)^2=31.22 \ m$
The location of the second stone at that time is
$y_2=\frac{1}{2}g(t_1-1.30 \ s)^2=\frac{1}{2}(9.80 \ m/s^2)(2.524 \ s-1.30 \ s)^2=7.34 \ s$
Thus the distance between the two stones is
$y_1-y_2=31.22 \ m - 7.34 \ m = 23.88 \ m$
Q#63
A person jumps off a diving board 4.0 m above the water’s surface into a deep pool. The person’s downward motion stops 2.0 m below the surface of the water. Estimate the average deceleration of the person while under the water.
Answer:
For the motion in the air, choose downward to be the positive direction and 0y = 0 to be at the height of the diving board. Then diver has $v_{0y}$ = 0 (assuming the diver does not jump upward or downward), $a_y$ = g = 9.80 $m/s^2$, and y = 4.0 m when reaching the surface of the water.
Find the diver’s speed at the water’s surface from Eq. $v_y^2=v_{0y}^2+2a_y(y-y_0) --> (*)$.
$v_y=\pm \sqrt{v_{0y}^2+2a_y(y-y_0)}$
$v_y=\pm \sqrt{0+2(9.80 \ m/s^2)(4.0 \ m)}=8.85 \ m/s$
For the motion in the water, again choose down to be positive, but redefine 0y = 0 to be at the surface of the water. For this motion, $v_{0y}$ = 8.85 m/s, v = 0, and $y-y_0$= 2.0 m.
Find the acceleration from Eq. (*), with x replaced by y.
$a_y=\frac{v_y^2-v_{0y}^2}{2(y-y_0)}$
$a_y=\frac{0^2-(8.85 \ m/s)^2}{2(2.0 \ m)}=-19.6 \ m/s^2$
The negative sign indicates that the acceleration is directed upward.
Q#64.
In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say 1.0 m long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. 2–47) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at 1.8 $m/s^2$ going downhill, and constantly at 2.6 $m/s^2$ going uphill. Suppose we have an uphill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup. Do the same for a downhill lie 7.0 m from the cup. What in your results suggests that the downhill putt is more difficult?
Answer:
First consider the “uphill lie,” in which the ball is being putted down the hill. Choose $x_0$ = 0 to be the ball’s original location and the direction of the ball’s travel as the positive direction.
The final velocity of the ball is v = 0, the acceleration of the ball is a = −1.8 $m/s^2$, and the displacement of the ball will be $x-x_0$ = 6.0 m for the first case and $x-x_0$ = 8.0 m for the second case.
Find the initial velocity of the ball from Eq. $v^2=v_0^2+2a(x-x_0)$, so
$v_0=\sqrt{v^2-2a(x-x_0)}$
$v_0=\sqrt{0-2(-1.8 m/s^2)(6.0 \ m)}=4.58 \ m/s$ and
$v_0=\sqrt{0-2(-1.8 m/s^2)(8.0 \ m)}=5.37 \ m/s$
The range of acceptable velocities for the uphill lie is 4.65 m/s to 5.37 m/s , a spread of 0.72 m/s. Now consider the “downhill lie,” in which the ball is being putted up the hill.
Use a very similar setup for the problem, with the basic difference being that the acceleration of the ball is now a = −2.6 $m/s^2$.
Find the initial velocity of the ball from Eq. $v^2=v_0^2+2a(x-x_0)$
$v_0=\sqrt{v^2-2a(x-x_0)}$
$v_0=\sqrt{0-2(-2.6 m/s^2)(6.0 \ m)}=5.59 \ m/s$ and
$v_0=\sqrt{0-2(-2.6 m/s^2)(8.0 \ m)}=6.45 \ m/s$
The range of acceptable velocities for the downhill lie is 5.59 m/s to 6.45 m/s, a spread of 0.86 m/s.
Because the range of acceptable velocities is smaller for putting down the hill, more control in putting is necessary, so putting the ball downhill (the “uphill lie”) is more difficult.
Q#65.
A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high (Fig. 2–48).
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?
Answer:
Choose upward to be the positive direction and $y_0$ = 0 to be at the throwing location of the stone.
The initial velocity is $v_{0y}$ = 15.5 m/s, the acceleration is a = −9.80 $m/s^2$, and the final location is y = −75 m.
(a) Using Eq. $y-y_0=v_{0y}t+\frac{1}{2}a_yt^2$ and substituting y for x, we have the following:
$-75 / m-0=(15.5 \ m/s)t+\frac{1}{2}(-9.80 \ m/s^2)t^2$
$(4.90 \ m/s^2)t^2-(15.5 \ m/s)t-75 \ m = 0$
$t_{1,2}=\frac{-(-15.5 \ m/s) \pm \sqrt{(-15.5 \ m/s)^2-4(4.90 \ m/s^2)(-75 \ m)}}{2(4.90 \ m/s^2)}$
$t_{1,2}=\frac{15.5 \pm \sqrt{240.25+1470}}{9.80}$
$t_1=\frac{15.5 + \sqrt{240.25+1470}}{9.80}=5.802 \ s$ or
$t_1=\frac{15.5 - \sqrt{240.25+1470}}{9.80}=-2.638 \ s$
The positive answer is the physical answer: t = 5.802 s
(b) Use Eq. $v_y=v_{0y}+a_yt$ to find the velocity just before hitting.
$v_y=15.5 \ m/s+(-9.80 \ m/s^2)(5.802 \ s)=-41.4 \ m/s$, so magnitude of velocity is $|v_y|=41.4 \ m/s$
(c) The total distance traveled will be the distance up plus the distance down. The distance down will be 75 m more than the distance up.
To find the distance up, use the fact that the speed at the top of the path will be 0. Then using Eq. $v_y^2=v_{0y}^2+2a_y(y-y_0)$ we have the following:
$y = y_0+\frac{v_y^2-v_{0y}^2}{2a_y}$
$y = 0+\frac{0-(15.5 \ m/s)^2}{2(-9.80 \ m/s^2)}=12.26 \ m$
Thus the distance up is 12.26 m, the distance down is 87.26 m, and the total distance traveled is 99.5 m.
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