problem 1 from rest, a car accelerated at 8m/s² for 10 seconds
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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Answer:
Problem 1:
a.) To find the final position of the car, we will use the equation:
s = ut + ½at²
Since the initial velocity (u) is 0 (car is at rest), the equation becomes:
s = 0 + ½at²
s = ½ * 8m/s² * (10s)²
s = 4 * 100m
s = 400m
So, the position of the car at the end of 10 seconds is 400 meters.
b.) To find the velocity of the car after 10 seconds, we'll use the equation:
v = u + at
v = 0 + 8m/s² * 10s
v = 80m/s
So the velocity of the car at the end of 10 seconds is 80 m/s.
Problem 2:
(Note: we need to convert the initial velocity from km/h to m/s)
20 km/h * (1000m/km) / (3600s/h) = 25/9 m/s
a.) s = ut + ½at²
s = (25/9 m/s) * 10s + ½ * 8m/s² * (10s)²
s = 250/3 m + 4 * 100m
s = 833.33m
So the position of the car at the end of 10 seconds is 833.33 meters.
b.) v = u + at
v = 25/9 m/s + 8m/s² * 10s
v = 80 + 25/9
v = 745/9 m/s
So the velocity of the car at the end of 10 seconds is 745/9 m/s (approximately 82.78 m/s).
Problem 3:
a.) Convert the final velocity from km/h to m/s.
72 km/h * (1000m/km) / (3600s/h) = 20 m/s
Next, find the acceleration using the formula: a = (v - u) / t
a = (20 m/s - 0) / 11.5s = 20/11.5 m/s² ≈ 1.739 m/s²
So the acceleration of the car is approximately 1.739 m/s².
b.) s = ut + ½at²
s = 0 + ½ * 1.739m/s² * (11.5s)²
s ≈ 0.87 * 132.25m
s ≈ 115.055m
So, the position of the car when it reaches 72 km/h (20 m/s) is approximately 115.055 meters.
Problem 4:
a.) Use the equation: v² = u² + 2as, solve for s (height of the building).
40m/s² = 20m/s² + 2*(-9.81m/s²)*s
s ≈ 30.56m
The height of the building is approximately 30.56 meters.
b.) Use the equation: v = u + at, solve for t (time in the air).
t = (v - u) / a = (40m/s - 20m/s) / -9.81m/s²
t ≈ 2.036s
So, the object was in the air for approximately 2.036 seconds.
Problem 5:
a.) Use the equation: v² = u² + 2as, solve for a (deceleration of the train).
0 = 40m/s² + 2a*100m
a = -40² /(2*100) = -1600 / 200 = -8 m/s²
The acceleration (deceleration) of the train is -8 m/s².
b.) Use the equation: v = u + at, solve for t (time it takes for the train to stop).
t = (v - u) / a = (0 - 40m/s) / -8m/s²
t = 5s
So, it takes 5 seconds for the train to stop.