Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.
So now we show it > 
Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.
Which the formula for constant acceleration would be > 
The initial velocity is 50mi/h 
When it stops the final velocity is 
Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27
Then we substitute the values in....
So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 
So this means that the car traveled in feet 70.8 ft before it came to a stop.
Solution :-
Given :
Distance 1 = 30 km
Distance 2 = 70 km
We know that speed = distance/time
and, Average speed = total distance/total time taken
When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour
Average speed = 9distance 1 + distance 2)/(time 1 + time 2)
AS time 2 or t2 is time taken for the second part of the journey of 70 km
⇒ 40 = 100/(1 + t2)
⇒ 40 + 40t2 = 100
⇒ 40t2 = 100 - 40
⇒ 40t2 = 60
⇒ t2 = 60/40
⇒ t2 = 1.5
So, t2 or time taken to travel the second part of the journey is 1.5 hours.
Speed of the second part of the journey = distance 2/time 2
⇒ 70/1.5
⇒ 46.666 km/hr or 46.7 km/hr.
Hence the answer is = 46.666 km/hr or 46.7 km/hr.
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Answer:
m₁ / m₂ = 1.3
Explanation:
We can work this problem with the moment, the system is formed by the two particles
The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero
p₀o = m₁ v₁ + m₂ v₂
pf = 0
m₁ v₁ + m₂ v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂= - (-6.2) / 4.7
m₁ / m₂ = 1.3
Another way to solve this exercise is to use the mass center relationship
Xcm = 1/M (m₁ x₁ + m₂ x₂)
We derive from time
Vcm = 1/M (m₁ v₁ + m₂v₂)
As they say the velocity of the center of zero masses
0 = 1/M (m₁ v₁ + m₂v₂)
m₁ v₁ + m₂v₂ = 0
m₁ / m₂ = -v₂ / v₁
m₁ / m₂ = 1.3
You get the net force acting on it ... the sum of the strengths and directions
of all the individual forces there may be.
