(a)We find the average speed = d/t = (100 m + 50 m)/[8.4 s + 1/3(8.4 s)] = 13.4 m/s.

(b) The displacement away from the trainer is 100 m - 50 m = 50 m; thus the average velocity is

For constant acceleration the average speed is 1/2(v + v0), thus

x = 1/2(v + v0)t: = 1/2(0 + 22.0 m/s)(5.00 s) = 55.0 m.

We use a coordinate system with the origin at the ground and up positive. When the package returns to the ground, its displacement is zero, so we have

y = y0 + v0 t + 1/2 a t^2;

0 = 115 m + (5.60 m/s)t + 1/2(-9.80 m/s^2)t^2.

The solutions of this quadratic equation are t = -4.31 s, and t = 5.44 s.

Because the package is released at t = 0, the positive answer is the physical answer: 5.44 s.

For the motion from A to B,

(a) The object is moving in the negative direction. The slope (the instantaneous velocity) is negative; the x-value is decreasing.

(b) Because the slope is becoming more negative (greater magnitude of the velocity), the object is speeding up.

(c) Because the velocity is becoming more negative, the acceleration is negative.

For the motion from D to E,

(d) The object is moving in the positive direction. The slope (the instantaneous velocity) is positive; the x-value is increasing.

(e) Because the slope is becoming more positive (greater magnitude of the velocity), the object is speeding up.

(f) Because the velocity is becoming more positive, the acceleration is positive.

(g) The position is constant, so the object is not moving, the velocity and the acceleration are zero.

. We use a coordinate system with the origin at the ground and up positive.

(a) We find the initial speed from the motion to the window:

v1^2 = v0^2 + 2a(y1 - y0); (14 m/s)^2 = v0^2 + 2(- 9.80 m/s^2)(25 m - 0), which gives v0 = 26 m/s.

(b) We find the maximum altitude from

v2^2 = v0^2 + 2a(y2 - y0);

0 = (26.2 m/s)^2 + 2(- 9.80 m/s^2)(y2 - 0), which gives y2 = 35 m.

(c) We find the time from the motion to the window:

v1 = v0 + a t1

14 m/s = 26.2 m/s + (- 9.80 m/s^2)t1 , which gives t1 = 1.2 s.

Thus it was thrown 1.2 s before passing the window.

(d) We find the time to reach the street from

y = y0 + v0 t + 1/2at^2;

0 = 0 + (26.2 m/s)t + 1/2(- 9.80 m/s^2)t^2.

This is a quadratic equation for t, which has the solutions t = 0 (the initial throw), 5.3 s.

Thus the time after the baseball passed the window is 5.3 s - 1.2 s = 4.1 s.

We use a coordinate system with the origin where the initial action takes place, as shown in the diagram.

The initial speed is (50 km/h)/(3.6 ks/h) = 13.9 m/s.

If she decides to stop, we find the minimum stopping distance from

v1^2 = v0^2 + 2a1(x1 - x0);

0 = (13.9 m/s)^2 + 2(- 6.0 m/s^2)x1 , which gives x1 = 16 m.

Because this is less than L1 , the distance to the intersection, she can safely stop in time.

If she decides to increase her speed, we find the acceleration from the time to go from 50 km/h to 70 km/h (19.4 m/s):

v = v0 + a2 t ;

19.4 m/s = 13.9 m/s + a2(6.0 s), which gives a2 = 0.917 m/s^2.

We find her location when the light turns red from

x2 = x0 + v0 t2 + 1/2 a2 t2^2 = 0 + (13.9 m/s)(2.0 s) + 1/2 (0.917 m/s^2)(2.0 s)^2 = 30 m.

Because this is L1, she is at the beginning of the intersection, but moving at high speed.

She should decide to stop!

If v_AG is the velocity of the automobile with respect to the ground, v_TG the velocity of the train with respect to the ground, and v_AT the velocity of the automobile with respect to the train, then v_AT = v_AG - v_TG

. If we use a coordinate system in the reference frame of the train with the origin at the back of the train, we find the time to pass from

x1 = v_AT t1 ; 1.10 km = (90 km/h - 80 km/h) t1 , which gives t1 = 0.11 h = 6.6 min.

With respect to the ground, the automobile will have traveled

x1_G = v_AG t1 = (90 km/h)(0.11 h) = 9.9 km.

If the automobile is traveling toward the train, we find the time to pass from x2 = v_AT t2 ;

1.10 km = [90 km/h - (- 80 km/h)]t2 , which gives t2 = 0.00647 h = 23.3 s.

With respect to the ground, the automobile will have traveled

x_2G = v_AG t2 = (90 km/h)(0.00647 h) =0.58 km.

We find the average acceleration from

v^2 = v0^2 + 2a(x2 - x1); (44 m/s)^2 = 0 + 2a(3.5 m), which gives a = 2.8x10^2 m/s^2.

We use a coordinate system with the origin at the ground and up positive.

(a) We find the velocity when the rocket runs out of fuel from v1^2 = v0^2 + 2a(y1 - y0);

v1^2 = 0+ 2(3.2 m/s^2)(1200 m - 0), which gives

v1 = 87.6 m/s = 88 m/s.

(b) We find the time to reach 1200 m from

v1 = v0 + a t1;

87.6 m/s = 0 + (3.2 m/s^2)t1, which gives t1 = 27.4 s = 27 s.

(c) After the rocket runs out of fuel, the acceleration is - g. We find the maximum altitude (where the velocity is zero) from

v2^2 = v1^2 + 2(- g)(h - y1);

0 = (87.6 m/s)^2 + 2(- 9.80 m/s2)(h - 1200 m), which gives h = 1590 m.

(d) We find the time from

v2 = v1 + (- g)(t2 - t1)

0 = 87.6 m/s + (- 9.80 m/s^2)(t2 - 27.4 s), which gives t2 = 36 s.

(e) We consider the motion after the rocket runs out of fuel:

v3^2 = v1^2 + 2(- g)(y3 - y1);

v3^2 = (87.6 m/s)^2 + 2(- 9.80 m/s^2)(0 - 1200 m), which gives v3 = - 177 m/s = - 1.8x10^2 m/s.

(f)We find the time from

v3 = v1 + (- g)(t3 - t1)

-177 m/s = 87.6 m/s + (- 9.80 m/s^2)(t3 - 27.4 s), which gives t3 = 54 s.

We use a coordinate system with the origin at the initial position of the front of the train. We can find the acceleration of the train from the motion up to the point where the front of the train passes the worker:

v1^2 = v0^2 + 2a(D - 0);

(20 m/s)^2 = 0 + 2a(180 m - 0),

which gives a = 1.11 m/s^2.

Now we consider the motion of the last car, which starts at - L, to the point where it passes the worker:

v2^2= v0^2 + 2a[D - (- L)]

= 0 + 2(1.11 m/s^2)(180 m + 90 m), which gives v2 = 24 m/s.

(a) If we assume that we are traveling at the speed limit, the time to pass through the farthest intersection is

t1 = (d1 + d2 + d3 + d4 + d5 + d6)/v1 = (10 m + 15 m + 50 m + 15 m + 70 m + 15 m)/(13.9 m/s) = 12.6 s.

Because this is less than the time while the lights are green, yes, you can make it through.

(b) We find the time required for the second car to reach the speed limit:

v_max = v0^2 + a2 t_2a;

13.9 m/s = 0 + (2.0 m/s^2)t_2a , which gives t_2a = 6.95 s.

In this time the second car will have traveled x_2a = v0^2 t1 + 1/2 a2 t_2a^2 = 0 + 1/2(2.0 m/s^2)(6.95 s)^2 = 48 m.

The time to travel the remaining distance at constant speed is t_2b = (d2 + d3 + d4 + d5 + d6 ? x_2a)/v_max

= (15 m + 50 m + 15 m + 70 m + 15 m - 48 m)/(13.9 m/s) = 8.42 s.

Thus the total time is

t_total = t_2a + t_2b = 6.95 s + 8.42 s = 15.4 s.

No, the second car will not clear all the lights.

For the vertical motion of James Bond we use a coordinate system with the origin at the ground and up positive. We can find the time for his fall to the level of the truck bed from y = y0 + v0 t + 1/2at^2;

1.5 m = 10 m + 0 + 1/2(- 9.80 m/s^2)t^2,

which gives t = 1.32 s.

During this time the distance the truck will travel is

x = x0 + v_truck t = 0 + (30 m/s)(1.32 s) = 39.6 m.

Because the poles are 20 m apart, he should jump when the truck is 2 poles away, assuming that there is a pole at the bridge.

Because there is no elapsed time when the light arrives, the sound travels one mile in 5 seconds. We find the speed of sound from

speed = d/t = (1 mi)(1610 m/1 mi)/(5 s) ~ 300 m/s.

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