Physics 140
HOMEWORK Chapter 2B
P. 40.
You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of v
0
= 55 km / hr; your best deceleration rate has the magnitude a = 5 .
18 m / s 2 . Your best reaction time to begin braking is T = 0 .
75 s. To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at 55 km/h if the distance to the intersection and the duration of the yellow light are (a) 40 m and 2.8 s, and (b) 32 m and 1.8 s? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).
———
The strategy to work this problem is to try both approaches for each of (a) and (b), and see whether either or both work. We note that dimensions are in meters, except v
0
, so convert that to m/s before we start.
v
0
= 55 km / hr = (55 km / hr) × (1000 m) / ( km) × (1 hr) / (3600 s) = 15 .
28 m / s
(a) Try continuing through the light: The distance you travel in 2.8 s is: d continue
= (15 .
28 m / s)(2 .
8 s) = 42 .
8 m.
continuing works.
Try braking. We will need separate calculations for the coast distance (reaction time, a = 0) and the braking distance ( a = − 5 .
18 m / s 2 ).
The distance you coast before reaction is: d coast
= v
0
T = (15 .
28 m / s)(0 .
75 s) = 11 .
5 m.
The distance to stop after you hit the brake is obtaining by rearranging 2 a ∆ x = ( v 2
∆ x = d brake
= ( v f
− v 2
0
) / 2 a = (0 − (15 .
28 m / s))
The total distance if you brake is:
2 / (2 · ( − 5 .
18 m / s 2 )) = 22 .
5 m.
− v 2
0
): d total
= 11 .
5 m + 22 .
5 m = 34 .
0 m.
Braking also works.
For part (a), EITHER strategy works.
(b) Try continuing: d continue
= (15 .
28 m / s)(1 .
8 s) = 27 .
5 m.
Continuing does NOT work, since the distance you needed was 32 m.
Try braking: Your d total
= 34 .
0 m, since the calculation is the same as for part (a).
Braking does
NOT work.
For part (b), NEITHER strategy works.
P. 43.
When a high-speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance
D = 676 m ahead (Fig. 2-29). The locomotive is moving at 29.0 km/h. The engineer of the highspeed train immediately applies the brakes. (a) What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided? (b) Assume that the engineer is at x = 0 when, at t = 0, he first spots the locomotive. Sketch x ( t ) curves for the locomotive and high-speed train for the cases in which a collision is just avoided and is not quite avoided.
———
Let’s do part (b) first; it will help with part (a):
(b) At t = 0, the locomotive is at x = 676 m and is moving at a constant velocity. So its plot x
L
( t ) is a straight line that starts at ( t = 0 , x = 676 m) and has a positive slope.
At t = 0, x
H
= 0 and it has a much steeper upward slope at t = 0. However, since the acceleration is negative, its continuation will be a parabola which is concave down.
For a collision, the line cuts the parabola, with the collision occuring at the first point of intersection. If they collide, something has to give, as they can’t continue on the same mathematical curves. However, suppose we put the trains on parallel tracks instead of the same track. In that case, the subsequent motion would follow the mathematical curves. The region where the parabola is above the line marks where the train is ahead of the locomotive; and the second intersection marks where the train has slowed enough that the locomotive moves back in front. If they barely miss, then the parabola curves away just fast enough that it barely misses the line.
(a) First, use v
L
= 29 km / hr = 8 .
056 m / s and v
0
= 161 km / hr = 44 .
722 m / s for the two speeds involved.

For a the vehicles to barely touch, the parabola just touches the line. Set up the equations for each vehicle: x
L
( t ) = D + v
L t constant velocity, a = 0 x
HST
= v
0 t + (1 / 2) a
0 t 2 a
0 is the acceleration, which is negative.
The concept might be to equate these expressions and solve for t , yielding the time t
0 at which they touch.
D + v
L t
0
= v
0 t
0
+ (1 / 2) at 2
0 rewrite in standard quadratic form:
(1 / 2) at 2
0
+ ( v
0
− v
L
) t
0
− D = 0.
The REALLY key realizations are (i) this is one equation in two unknowns, so we don’t expect to solve for both; and (ii) the condition for barely touching is that the quadratic have only one real solution (a double root). That means the discriminant is zero:
( v
0
− v
L
) 2 − 4(1 / 2) a ( − D ) = 0.
solve for a : a = − ( v
0
− v
L
) 2 / 2 D = − (44 .
72 − 8 .
06) 2 / (2 · (676 m) = − 0 .
994 m / s 2 .
Since the problem stated “magnitude of deceleration,” take the absolute value: a = | ~a | = 0 .
994 ms 2 .
This is plotted in a separate pdf; the line is the slow locomotive; the black parabola is the curve to barely touch (solution); the red curve shows the results of a few percent too little deceleration; and the green curve shows a few percent more than the absolute minimum.
P. 47.
At a construction site a pipe wrench struck the ground with a speed of 24 m/s. (a) From what height was it inadvertently dropped? (b) How long was it falling? (c) Sketch graphs of y , v , and a versus t for the wrench.
———
(a) Make the + y axis upward, and put y = 0 at ground level. Then a = − g = − 9 .
8 m / s 2 acceleration equations are
. The constant v = v
0
− gt and ∆ y = v
0 t − (1 / 2) gt 2
Since the wrench was “dropped,” v
0
= 0.
and v 2 − v
0
2 = 2 a ∆ y .
Use the time-free equation with
∆ y = ( v 2 − v 2
0
So the height was 29 .
4 m .
a y
) / ( − 2 g ) = (24 m / s)
= − g :
2 / ((2 · ( − 9 .
8 m / s 2 )) = − 29 .
4 m.
(b) Solve the velocity equation for the time when v = − 24 m / s: t hit
= (24 m / s) / (9 .
8 m / s 2 ) = 2 .
45 s
(c) y ( t ) is a parabola whose apex is (0 , 29 .
4 m) and which intersects the x -axis at (2 .
45 s , 0).
v ( t ) is a line through (0 , 0) and a slope of − 9 .
8 m / s 2 .
a ( t ) is a constant function (horizontal line) at a = − 9 .
8 m / s 2 .
P. 49.
A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?
———
(a) ∆ y = v
0 y t + (1 / 2) a y t 2 . Use t
0 for the time it hits ground. Substitute known values ∆ y = − 80 m; v
0
= +12 m
−
4 .
9 t 2
0
/ s; and
80 m = (+12 m a
/ y s)
= − g : t
0
− (1 / 2)(9 .
8 m /
− 12 t
0
− 80 = 0 ⇒ t
0
= (12 ±
(b) v = v
0
− 9 .
8 t
0
= 12 m / s − (9 .
8 m / s 2 s 2
√
)(5 t
144 + 1568)
.
2
0
. Suppress units and put in standard form:
45 s) = −
/
41 .
9 .
8 =
4 m /
5 s.
.
45 s .
(The − 3 .
00 s solution represents extending the trajectory backwards in time.)
Speed = 41 .
4 m / s .
P. 53.
A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity, that is 12 m from the point of impact when the key is released. What is the speed of the boat?
———
A compound problem. First, calculate the time the key takes to drop. After that, it’s a snap to figure how fast the model boat was moving.
∆ y = v
0 t − (1 / 2) gt 2 = − (1 / 2) gt 2 v boat
= (12 m) / (3 .
03 s) = 3 .
96 m / s .
⇒ t = p − 2∆ y/g = p ( − 2)( − 45 m) / (9 .
8 m / s 2 ) = 3 .
03 s

P. 69.
How far does the runner whose velocity-time graph is shown in Fig. 2-37 travel in 16 s? The
Figure’s vertical scaling is set by v s
= 8 .
0 m / s.
———
The net distance traveled is simply the area under the curve. If we use calculus, the integrals are trivial:
R t dt and R dt . It’s messy to set up the linear equations, though. We can also get the area by geometry: areas of triangles ( A = BH/ 2), trapezoids ( A = ( B
1
+ B
2
)( H ) / 2), and rectangles.
Interval (s)
0 to 2
2 to 10
10 to 12
12 to 16
Figure
Triangle
Rectangle
Trapezoid
Rectangle
Distance(m)
8
64
12
16
Total 100 m
Note that the “area under the curve” comes out in a linear distance, as indicated by the product of the horizontal axis units (s) and the vertical axis units (m/s).