A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary
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Answer:
The graphs are attached
Explanation:
We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.
At constant velocity, v = distance/time
time(t) = distance(d)/velocity(v)
t1 = 100/25
t1 = 4 s
Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.
It means, he decelerate and final velocity is zero.
Thus;
v² = u² + 2as
0² = 25² + 2a(50)
25² = - 100a
625 = - 100a
a = - 625/100
a = - 6.25 m/s²
v = u + at
0 = 25 + (-6.25t)
25 = 6.25t
t = 25/6.25
t = 4 s
With the values gotten, kindly find attached the distance-time and velocity-time graphs.
