A car starts out traveling at 35 m/s. The car hits the brakes and decelerates at a rate of 3 m/s^2 for 5 seconds. What Distance
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The second stone hits the ground exactly one second after the first.
The distance traveled by each stone down the cliff is calculated using second kinematic equation;
where;
- <em>t is the time of motion </em>
- <em />
<em> is the initial vertical velocity of the stone = 0</em>
The time taken by the first stone to hit the ground is calculated as;
When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as
Thus, we can conclude that the second stone hits the ground exactly one second after the first.
"<em>Your question is not complete, it seems be missing the following information;"</em>
A. The second stone hits the ground exactly one second after the first.
B. The second stone hits the ground less than one second after the first
C. The second stone hits the ground more than one second after the first.
D. The second stone hits the ground at the same time as the first.
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1) Velocity: 9.9 m/s and 14 m/s
The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:
where
v is the final velocity
u = 0 is the initial velocity (the diver starts from rest)
is the acceleration of gravity
s is the displacement
For the diver jumping from 5 m, s = 5 m, so
For the diver jumping from 10 m, s = 10 m, so
2) Time: 1.01 s and 1.43 s
The time of flight of each diver can be found by using the other suvat equation
And since u = 0, it can be reduced to
For the diver jumping from 5 m, s = 5 m, so we find
For the diver jumping from 10 m, s = 10 m, so we find