At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.1 m/s2. At the same instant a truck, traveling with a constant speed of 9.5 m/s, overtakes and passes the automobile.
(a) How far beyond the traffic signal will the automobile overtake the truck?
(b) How fast will the car be traveling at that instant?
Thursday, January 30, 2014
The brakes on your car can slow you at a rate of 5.2 m/s2. (a) If you are going 137 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 90 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph x versus t and v versus t for such a slowing.
The brakes on your car can slow you at a rate of 5.2 m/s2. (a)
If you are going 137 km/h and suddenly see a state trooper, what is
the minimum time in which you can get your car under the 90
km/h speed limit? (The answer reveals the futility of braking to
keep your high speed from being detected with a radar or laser
gun.)
We choose the positive direction to be that of the initial velocity of the car
(implying that a < 0 since it is slowing down). We assume the acceleration is constant
and use Table 2-1.
(a) Substituting v0 = 137 km/h = 38.1 m/s,
v = 90 km/h = 25 m/s,
to covert km/h to meter/second multiply the 137 & 90 by 0.27778
and
a = –5.2 m/s2 (from the question)
into v = v0 + at, we obtain
t = (25-38.1)/-5.2= 2.5s
If you are going 137 km/h and suddenly see a state trooper, what is
the minimum time in which you can get your car under the 90
km/h speed limit? (The answer reveals the futility of braking to
keep your high speed from being detected with a radar or laser
gun.)
We choose the positive direction to be that of the initial velocity of the car
(implying that a < 0 since it is slowing down). We assume the acceleration is constant
and use Table 2-1.
(a) Substituting v0 = 137 km/h = 38.1 m/s,
v = 90 km/h = 25 m/s,
to covert km/h to meter/second multiply the 137 & 90 by 0.27778
and
a = –5.2 m/s2 (from the question)
into v = v0 + at, we obtain
t = (25-38.1)/-5.2= 2.5s
If the position of a particle is given by x = 15t - 1t3, where x is in meters and t is in seconds, answer the following questions.
If the position of a particle is given by x = 15t - 1t3, where x is in meters and t is in seconds, answer the following questions.
(a) When, if ever, is the particle's velocity zero? (Enter the value of t or 'never'.)
(b) When is its acceleration a zero?
(a) When, if ever, is the particle's velocity zero? (Enter the value of t or 'never'.)
(b) When is its acceleration a zero?
The position of a particle moving along the x axis is given in centimeters by x = 7.00 + 2.50t 3, where t is in seconds. Consider the time interval t = 2.00 s to t = 3.00 s.
The position of a particle moving along the x axis is given in centimeters by x = 7.00 + 2.50t 3, where t is in seconds. Consider the time interval t = 2.00 s to t = 3.00 s.
(a) Calculate the average velocity.
(b) Calculate the instantaneous velocity at t = 2.00 s.
(c) Calculate the instantaneous velocity at t = 3.00 s.
(d) Calculate the instantaneous velocity at t = 2.50 s.
(e) Calculate the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s.
(a) Calculate the average velocity.
(b) Calculate the instantaneous velocity at t = 2.00 s.
(c) Calculate the instantaneous velocity at t = 3.00 s.
(d) Calculate the instantaneous velocity at t = 2.50 s.
(e) Calculate the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s.
The position of an object moving in a straight line is given by x = 7t - 5 t 2 + 3t 3, where x is in meters and t in seconds.
The position of an object moving in a straight line is given by x = 7t - 5 t 2 + 3t 3, where x is in meters and t in seconds.
(a) What is the position of the object at t = 1s?
What is the position of the object at t = 2?
What is the position of the object at t = 3?
What is the position of the object at t = 4 s?
(b) What is the object's displacement between t = 0 and t = 4 s?
(c) What is the average velocity for the time interval from t = 2 s to t = 4 s?
(a) What is the position of the object at t = 1s?
What is the position of the object at t = 2?
What is the position of the object at t = 3?
What is the position of the object at t = 4 s?
(b) What is the object's displacement between t = 0 and t = 4 s?
(c) What is the average velocity for the time interval from t = 2 s to t = 4 s?
Tuesday, January 21, 2014
Prior to adopting metric systems of measurement, the United Kingdom employed some challenging measures of liquid volume. A few are shown in the table (below). (a) Complete the table.
Prior to adopting metric systems of measurement, the United Kingdom employed some challenging measures of liquid volume. A few are shown in the table (below).
(a) Complete the table.
(b) The volume of 1 bag is equivalent to a volume of 0.1091 m3. If an old British story has a witch cooking up some vile liquid in a cauldron with a volume of 2.1 cauldrons, what is the volume in terms of cubic meters?
(b) The volume of 1 bag is equivalent to a volume of 0.1091 m3. If an old British story has a witch cooking up some vile liquid in a cauldron with a volume of 2.1 cauldrons, what is the volume in terms of cubic meters?
Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumps 2.5 in. of rain in 30 min on a town of area 44 km2. What volume of water, in acre-feet, fell on the town?|
Hydraulic engineers often use, as a unit of volume of water, the
"acre-foot", defined as the volume of water that will cover 1 acre of
land to a depth of 1 ft. A severe thunderstorm dumps 2.5 in. of rain in 30 min on a town of area 44 km2. What volume of water, in acre-feet, fell on the town?|
A "gry" is an old English measure for length, defined as 1/10 of a line, where "line" is another old English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a "point", defined as 1/72 inch. What is an area of 0.55 gry2 in terms of points squared (points2)?
A "gry" is an old English measure for length, defined as 1/10 of a line,
where "line" is another old English measure for length, defined as 1/12
inch. A common measure for length in the publishing business is a
"point", defined as 1/72 inch. What is an area of 0.55 gry2 in terms of points squared (points2)?
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