Response:
0.60 m/s
Details:
The average speed between times t = a and t = b can be expressed as:
v_avg = (x(b) − x(a)) / (b − a)
Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:
v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)
v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0
v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0
v_avg = [0.96 − (-0.84)] / 3.0
v_avg = 0.60
The average speed calculated is 0.60 m/s.
Answer:
All observers are accurate.
Explanation:
This situation reflects a matter of reference frames regarding the book's motion as perceived by different observers.
From their distinct frames of reference, each observer's perspective is valid.
Observer A is in an inertial reference frame.
Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.
Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.
Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.
This is exemplified by observer B, who notes the car's force impacting the book's velocity.
Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.
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Answer:
1/7 kg
Explanation:
Refer to the attached diagram for enhanced clarity regarding the question.
One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.
g denotes the acceleration due to gravity.
Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.
Given M = 1.0 kg and a = 3/4g.
By applying Newton's second law; 
For the body with mass m;
T - mg = ma... (1)
For the body with mass M;
Mg - T = Ma... (2)
Combining equations 1 and 2 gives;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into this equation leads to;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
Multiplying gives: 28m = 4
m = 1/7 kg
Hence, the mass of the other box is 1/7 kg
1 hour = 3,600 seconds
1 km = 1,000 meters
75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s
The mean speed during the deceleration is
(1/2)(20-5/6 + 0) = 10-5/12 m/s.
Traveling at this average speed for 21 seconds,
the bus covers
(10-5/12) × (21) = 218.75 meters.