A runner is at the starting line and hears the starting gun. He starts running from rest with a constant acceleration ai = 0.53 m/s2. He crosses the finish line at d = 100 m and then begins slowing down. It takes him a time interval tr to cross the finish line. It takes him a time interval ts = 7.5 s to return to rest after crossing the finish line. Use the given coordinate system where x is positive to the right. a)Enter an expression for the time, tr, it takes for the runner to cross the finish line, in terms of d and ai
Answers
To find the time, tr, it takes for the runner to cross the finish line, we can use one of the kinematic equations that relate displacement, initial velocity, acceleration, and time. Since the runner starts from rest, his initial velocity is zero, so we can use the equation:
Δx=v0t+21at2
where Δx is the displacement, v0 is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values of Δx=d, v0=0, and a=ai, we get:
d=0+21ait2
Solving for t, we obtain:
t=ai2d
This is the expression for the time, tr, it takes for the runner to cross the finish line, in terms of d and ai.
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For example, a hydrogen atom forms one bond, an oxygen atom forms two, and carbon forms four bonds. Look at that molecule of water again - each hydrogen has one bond, and the oxygen in the middle has two bonds. Molecules can be much bigger.
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Take them one at a time:
First piece:
From rest, accelerates 1.8 m/s² for 15 sec.
Speed at the end of the 15 sec = 1.8 x 15 = 27 m/s
Distance covered = (average speed) x (time) = (13.5) x (15) = 202.5 meters
Middle piece:
Starts this piece at 27 m/s.
Runs at constant speed for 35 sec.
Covers (27 x 35) = 945 meters
Finishes this piece (15 + 35) = 50 seconds after leaving the first station
Finishes this piece (202.5 + 945) = 1147.5 meters from the first station.
Last piece:
Starts this piece 50 seconds after leaving the first station,
1147.5 meters from it, still moving at 27 m/s.
Decelerates at 2.0 m/s² .
--> It takes 27/2.0 = 13.5 seconds to come to a stop.
--> Distance covered during this piece = (average speed) x (time)
= (13.5 m/s) x (13.5 sec) = 182.25 meters
Finishes this piece (50 + 13.5) = 63.5 seconds after leaving the first station
Finishes this piece (1147.5 + 182.25) = 1,329.75 m from the first station
Distance between the stations (along the track) = 1,329.75 meters.
Average speed = (total distance) / (total time) = 1,329.75 / 63.5 = 20.941 m/s.
I sure hope I didn't make any stupid arithmetic blunders here,
because I really don't feel like going back and checking it.
Let me know, and I'll fix them.
Answers
1. How fast is the blue car going 1.8 seconds after it starts?
Recall this kinematic equation:
Vf = Vi + aΔt
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest)
a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)
Δt = 1.8 s
Substitute the terms in the equation with the given values and solve for Vf:
Vf = 0 + 3.7×1.8
Vf = 6.66 m/s
2. How fast is the blue car going 10.0 seconds after it starts?
The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.
We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:
Vf = 0 + 3.7×4.4
Vf = 16.28 m/s
The constant velocity at which the car is traveling at t = 10s is 16.28 m/s
3. How far does the blue car travel before its brakes are applied to slow down?
We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.
Traveling under constant acceleration:
Recall this kinematic equation:
d = ×Δt
d is the distance traveled.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 16.28 m/s (determined from question 2).
Δt = 4.4 s
Substitute the terms in the equation with the given values and solve for d:
d = ×4.4
d = 35.8 m
Traveling at constant velocity:
Recall the relationship between velocity and distance:
d = vΔt
d is the distance traveled.
v is the velocity.
Δt is the amount of elapsed time.
Given values:
v = 16.28 m/s (the constant velocity from question 2).
Δt = 8.3 s (the time interval during which the car travels at constant velocity)
Substitute the terms in the equation with the given values:
d = 16.28×8.3
d = 135.1 m
Add up the distances traveled.
d = 35.8 + 135.1
d = 170.9 m
4. What is the acceleration of the blue car once the brakes are applied?
Recall this kinematic equation:
Vf²=Vi²+2ad
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration
d is the distance traveled.
Given values:
Vi = 16.28 m/s
Vf = 0 m/s
d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)
Substitute the terms in the equation with the given values and solve for a:
0² = 16.28²+2a×45.1
a = -2.94 m/s²
5. What is the total time the blue car is moving?
We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.
Recall again:
d = ×Δt
Given values:
d = 45.1 m
Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).
Vf = 0 m/s (the car slows to a stop).
Substitute the terms in the equation with the given values and solve for Δt:
45.1 = ×Δt
Δt = 5.54s
Add up the times to get the total travel time:
t = 4.4 + 8.3 + 5.54 =
t = 18.24s
6. What is the acceleration of the yellow car?
Recall this kinematic equation:
d = ViΔt + 0.5aΔt²
d is the distance traveled.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
d = 216 m (both cars meet at 216m)
Vi = 0 m/s (the car starts at rest)
Δt = 18.24 s (take the same amount of time to reach 216m)
Substitute the terms in the equation with the given values and solve for a:
216 = 0×18.24 + 0.5a×18.24²
a = 1.3 m/s²
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Answers
Not so ! Not only is that statement essentially false,
but it involves significant prevarication as well, too.
Mass is the measure of the quantity of matter in an object.
Volume is a measure of the quantity of space filled by the
object.
The same mass can have many different volumes, usually
depending on its temperature. Just consider one kilogram
of ice, water, and steam . . .
B2+C2-->2BC
A+BC-->AB+C
Exothermic
AB+C-->AC+B
A2+C2-->2AC
A2+B2-->2AB
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Endothermic
B2+C2-->2BC
A2+C2-->2AC
A2+B2-->2AB
Exothermic
AB+C-->AC+B
A+BC-->AB+C
Answer:
The correct answer is
Endothermic:
AB+C-->AC+B
A2+C2-->2AC
B2+C2-->2BC
Exothermic:
A+BC-->AB+C
A2+B2-->2AB
Explanation:
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