problem 1 from rest, a car accelerated at 8m/s² for 10 seconds to
a.) what is the position of the car at the end of 10 seconds?
b.) what is the vilocity of the car at the end of 10 seconds
problem 2 what an initial velocity of 20 km/h, a car accelerated at 8m/s for 10 seconds.
a.) what is the position of the car at the end of 10 seconds
b.) what is the velocity of the car at the end of 10 seconds
problem 3 a car accelerates uniformly from 0 to 72 km/s in 11.5 seconds.
a.) what is the acceleration of the car in m/s²?
b.) what is the position of the car by the time it reaches the velocity of 72 km/s?
proble 4 an object is thrown straight down from the top of a building at a speen of 20m/s. it hits the groud with a speed of 40m/s
a.) how high is the building?
b.)how long was the object in the earth
problem 5 a train brakes from 40m/s to a stop over a distance of 100m.
a.) what is the acceleration of the train?
b.) how much time does it take the train to stop?
a.) what is the position of the car at the end of 10 seconds?
b.) what is the vilocity of the car at the end of 10 seconds
problem 2 what an initial velocity of 20 km/h, a car accelerated at 8m/s for 10 seconds.
a.) what is the position of the car at the end of 10 seconds
b.) what is the velocity of the car at the end of 10 seconds
problem 3 a car accelerates uniformly from 0 to 72 km/s in 11.5 seconds.
a.) what is the acceleration of the car in m/s²?
b.) what is the position of the car by the time it reaches the velocity of 72 km/s?
proble 4 an object is thrown straight down from the top of a building at a speen of 20m/s. it hits the groud with a speed of 40m/s
a.) how high is the building?
b.)how long was the object in the earth
problem 5 a train brakes from 40m/s to a stop over a distance of 100m.
a.) what is the acceleration of the train?
b.) how much time does it take the train to stop?
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Problem 1:
a.) The position of the car can be calculated using the formula:
s = ut + 1/2at²
Where s is the position, u is the initial velocity (which is 0 in this case), a is the acceleration (which is 8m/s²), and t is the time (which is 10 seconds). Plugging in the values we get:
s = 0 + 1/2(8)(10)²
s = 400m
Therefore, the position of the car at the end of 10 seconds is 400 meters.
b.) The velocity of the car at the end of 10 seconds can be calculated using the formula:
v = u + at
Where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration (which is 8m/s²), and t is the time (which is 10 seconds). Plugging in the values we get:
v = 0 + (8)(10)
v = 80m/s
Therefore, the velocity of the car at the end of 10 seconds is 80 meters per second.
Problem 2:
a.) The position of the car can be calculated using the formula:
s = ut + 1/2at²
Where s is the position, u is the initial velocity (which is 20 km/h = 5.56 m/s in this case), a is the acceleration (which is 8m/s²), and t is the time (which is 10 seconds). Plugging in the values we get:
s = (5.56)(10) + 1/2(8)(10)²
s = 278m
Therefore, the position of the car at the end of 10 seconds is 278 meters.
b.) The velocity of the car at the end of 10 seconds can be calculated using the formula:
v = u + at
Where v is the final velocity, u is the initial velocity (which is 20 km/h = 5.56 m/s in this case), a is the acceleration (which is 8m/s²), and t is the time (which is 10 seconds). Plugging in the values we get:
v = 5.56 + (8)(10)
v = 85.56m/s
Therefore, the velocity of the car at the end of 10 seconds is 85.56 meters per second.
Problem 3:
a.) The acceleration of the car can be calculated using the formula:
a = (v-u)/t
Where a is the acceleration, v is the final velocity (which is 72 km/s = 72,000 m/s in this case), u is the initial velocity (which is 0 in this case), and t is the time (which is 11.5 seconds). Plugging in the values we get:
a = (72,000 - 0)/11.5
a = 6,260.87m/s² (rounded to 3 decimal places)
Therefore, the acceleration of the car is 6,260.87 meters per second squared.
b.) The position of the car can be calculated using the formula:
s = ut + 1/2at²
Where s is the position, u is the initial velocity (which is 0 in this case), a is the acceleration (which is 6,260.87m/s²), and t is the time (which is 11.5 seconds). Plugging in the values we get:
s = 0 + 1/2(6,260.87)(11.5)²
s = 4,219,565.22m
Therefore, the position of the car by the time it reaches the velocity of 72 km/s is 4,219,565.22 meters.
Problem 4:
a.) The height of the building can be calculated using the formula:
s = ut + 1/2at²
Where s is the height of the building, u is the initial velocity (which is 20m/s), a is the acceleration due to gravity (which is -9.81m/s²), and t is the time it takes for the object to hit the ground (which we need to find). We know that the final velocity of the object is 40m/s. Using the formula:
v² = u² + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the height of the building. Plugging in the values we get:
40² = 20² + 2(-9.81)s
s = 78.4m
Therefore, the height of the building is 78.4 meters.
b.) The time it takes for the object to hit the ground can be calculated using the formula:
v = u + at
Where v is the final velocity (which is 40m/s), u is the initial velocity (which is 20m/s), a is the acceleration due to gravity (which is -9.81m/s²), and t is the time it takes for the object to hit the ground. Plugging in the values we get:
40 = 20 + (-9.81)t
t = 2.04s
Therefore, the object was in the air for 2.04 seconds.
Problem 5:
a.) The acceleration of the train can be calculated using the formula:
a = (v-u)/t
Where a is the acceleration, v is the final velocity (which is 0 in this case), u is the initial velocity (which is 40m/s in this case), and t is the time it takes for the train to stop (which we need to find). We know that the distance the train travels before stopping is 100m. Using the formula:
v² = u² + 2as
Where v is the final velocity (which is 0), u is the initial velocity (which is 40m/s), a is the acceleration, and s is the distance traveled by the train before stopping (which is 100m). Plugging in the values we get:
0² = 40² + 2a(100)
a = -8m/s²
Therefore, the acceleration of the train is -8 meters per second squared.
b.) The time it takes for the train to stop can be calculated using the formula:
v = u + at
Where v is the final velocity (which is 0), u is the initial velocity (which is 40m/s), a is the acceleration (which is -8m/s²), and t is the time it takes for the train to stop. Plugging in the values we get:
0 = 40 + (-8)t
t = 5s
Therefore, it takes the train 5 seconds to stop.