Hello people~
2 answers:
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Ok i apologise for the messy working but I'll try and explain my attempt at logic
Also note i ignore any air resistance for this.
First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.
Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.
I also noted the "initial" and "final" height of the swing with hi and hf respectively.
So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.
So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).
Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.
The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).
I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx
Answer:
v = 1.98 mph
Explanation:
Given that,
Speed to travel one mile is 100 mph
Speed to travel another mile is 1 mph
The formula used to find your average speed is given by :
Putting the values, we get :
v = 1.98 mph
So, yours average speed is 1.98 mph.
Answer:
Part 1) Time of travel equals 61 seconds
Part 2) Maximum speed equals 39.66 m/s.
Explanation:
The final speed of the train when it completes half of it's journey is given by third equation of kinematics as
where
'v' is the final speed
'u' is initial speed
'a' is acceleration of the body
's' is the distance covered
Applying the given values we get
Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as
Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance
Thus total time of journey equals
Part b)
the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals