Answer:
48 m
Explanation:
Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.
The eastbound train
First convert km/h to m/s
(97 × 1000)/3600
97000/3600
26.944444 m/s
As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion
V^2 = U^2 - 2as
O = 26.944^2 - 2 × 3.5 S
726 = 7S
S = 726/7
S1 = 103.7 m
The westbound train
Convert km/h to m/s
(127×1000)/3600
127000/3600
35.2778 m/s
Using third equation of motion
V^2 = U^2 - 2as
0 = 35.2778^2 - 2 × 4.2 × S
1244.52 = 8.4S
S = 1244.52/8.4
S2 = 148.2 m
S1 + S2 = 103.7 + 148.2 = 251.86
The distance between them once they stop will be
300 - 251.86 = 48.14 m
Therefore, the distance between them once they stop is 48 metres approximately.
Answer:
1.67m
5m/s
Explanation:
Given parameters:
Wavelength of the wave = 3m
Speed of the wave = 5m/s
Unknown:
Frequency of the wave = ?
Speed of the water waves = ?
Solution:
The distance between crest and the adjacent trough of water waves is known as the wavelength of a wave.
To find the frequency ;
V = f∧
V is the speed of the wave
f is the frequency
∧ is the wavelength
Insert the parameters and find the frequency;
f = V/ ∧ = 5 / 3 = 1.67Hz
The rate at which the wave passed a given point is the speed of the wave and it is 5m/s
