Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.7 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.3 seconds, before applying the brakes such that the carâs speed decreases uniformly coming to rest 216.0 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop. 1. How fast is the blue car going 1.8 seconds after it starts? 2. How fast is the blue car going 10.0 seconds after it starts? 3. How far does the blue car travel before its brakes are applied to slow down? 4. What is the acceleration of the blue car once the brakes are applied? 5. What is the total time the blue car is moving? 6. What is the acceleration of the yellow car?
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
Answer:
s(t) = (t1 - t2)* [((1/t1 - 3) + (1/t0 - 3))/2]
Explanation:
We will assume that v(t) is in units of m/s and t (time) is in seconds.
v(t)=1/t−3
At time 0 (initial) the equation tells us the particle has a velocity of
v(t)=1/t−3
v(0)=1/(0)−3
v(0) = - 3 m/s
The particle is moving from right to left (the negative sign) at a rate of 3 m/s.
The position of the particle would be the average velocity times the time traveled.
Distance = Velocity x Time (with Velocity being the average between times t0 and t1)
We'll use s(t) for displacement for time t.
s(t) = v*t
We need the average velocity for the time period t0 to t1.
Let t0 and t1 be the initial and final times in which the measurement takes place.
At time t0 the velocity is = 1/t0 - 3
At time t1 the velocity is = 1/t1 - 3
The displacement is the average velocity between the two points, t0 and t1. This can be written as:
[(1/t1 - 3) + (1/t0 - 3)/\]/2
Displacement: s(t) = (t1 - t2)* [((1/t1 - 3) + (1/t0 - 3))/2]
Answers
Answer:
retains a fixed volume and shape
rigid - particles locked into place
not easily compressible
little free space between particles
does not flow easily
rigid - particles cannot move/slide past one another
Explanation:
1. Behavior: Solids have a definite shape and volume.
Energy: Solids possess potential energy due to the arrangement of their particles.
2. Behavior: Solids are rigid and maintain their shape when subjected to external forces.
Energy: Solids exhibit elastic potential energy when deformed and can store mechanical energy.
3. Behavior: Solids have a fixed melting point at which they transition to a liquid state.
Energy: Solids require thermal energy (heat) to overcome intermolecular forces and transition to a liquid state.
B. avoid a post-trip inspection
C. know how far you must travel
D. save money on fuel
Answers
Answer:
A. find problems that could cause a collision or breakdown
Explanation:
A pretrip inspection allows the driver to drive safely because he feels satisfied since he already had checked each and everything in the beginning. This increases the focus on driving. In addition to it there are less chances of happenings and technical faults in the vehicle. It also plays a vital role in the safety circumstances as well.
A.To find problems that could cause a collision or breakdown.
Explanation
A pre-trip inspection is the trip made by the organizers before the due date of the trip.
There are many reasons why pre-trip inspection is necessary. Some of them include:
1. Finding out how much information is available,
2. The cost to be incurred,
3. Assessing the route and the safety of the place,
4. The most important been to help in preparation of the budget and many others
From the choices given, the most appropriate answer is A. To find problems that could cause a collision or breakdown.
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Answers
Answer:
The correct answer is Joseph Stalin.
Explanation:
Joseph Vissarionovich Stalin was a Soviet politician who led the soviet union from 1929 - 1953. Stalin self declared himself as the dictator by 1930s. He was the person behind the Great Purge or "Great Terror" in which around a million were imprisoned and some 750,000 were executed. This campaign was led just to execute all those who were a threat to his party.