Starting from rest, a boulder rolls down a hill with constant acceleration and travels 2.00m duering the first second.
How far does it travel during the second second?
How fast is it moving at the end of the first second? At the end of the second second?
To solve this problem, we need to use the equations of motion for constant acceleration.
1. How far does it travel during the second second?
The formula we can use to solve this is:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
initial velocity = 0 (as it starts from rest)
time = 2 seconds
acceleration = constant
Since the acceleration remains constant, we can use the formula:
distance = (1/2) * acceleration * time^2
Plugging in the values:
distance = (1/2) * acceleration * (2)^2
2m = 2 * acceleration
Therefore, the boulder travels 2 meters during the second second.
2. How fast is it moving at the end of the first second? At the end of the second second?
The formulas we can use to solve these are:
velocity = initial velocity + acceleration * time
Given:
initial velocity = 0 (as it starts from rest)
time = 1 second for the end of the first second
time = 2 seconds for the end of the second second
acceleration = constant
Plugging in the values for the first second:
velocity = 0 + acceleration * 1
velocity = acceleration
Therefore, at the end of the first second, the boulder is moving at a velocity equal to the acceleration.
Plugging in the values for the second second:
velocity = 0 + acceleration * 2
velocity = 2 * acceleration
Therefore, at the end of the second second, the boulder is moving at a velocity equal to twice the acceleration.
To find out how far the boulder travels during the second second, we can use a kinematic equation.
First, we need to determine the acceleration. Because the boulder rolls down the hill with constant acceleration, the acceleration during the first second will be the same as the acceleration during the second second. Let's call this acceleration "a."
Given information:
Initial velocity (u) = 0 (starting from rest)
Distance traveled during the first second (s1) = 2.00 m
Time duration (t1) = 1.0 s
We can use the equation: s = ut + 0.5at^2
Plugging in the values for the first second:
2.00 m = 0 + 0.5a(1.0 s)^2
Simplifying the equation:
2.00 m = 0.5a
Now, let's find the value of "a":
2.00 m = 0.5a
a = (2.00 m) / (0.5)
a = 4.00 m/s^2
Since the acceleration remains constant, the acceleration during the second second is also 4.00 m/s^2.
Now, to find out how far the boulder travels during the second second (s2), we can use the same kinematic equation:
s2 = ut + 0.5at^2
Given information for the second second:
Initial velocity (u) = 0 (since it starts from rest again)
Time duration (t2) = 1.0 s (since we are looking at the second second)
s2 = 0 + 0.5(4.00 m/s^2)(1.0 s)^2
s2 = 0 + 0.5(4.00 m/s^2)(1.0 s)
s2 = 0 + 2.00 m
s2 = 2.00 m
Therefore, the boulder travels 2.00 meters during the second second.
To find out how fast the boulder is moving at the end of the first second and the end of the second second, we can use the equation: v = u + at
For the first second:
Initial velocity (u) = 0 (starting from rest)
Acceleration (a) = 4.00 m/s^2 (calculated earlier)
Time duration (t) = 1.0 s
v1 = 0 + (4.00 m/s^2)(1.0 s)
v1 = 0 + 4.00 m/s
v1 = 4.00 m/s
At the end of the first second, the boulder is moving at a speed of 4.00 m/s.
For the second second:
Initial velocity (u) = 0 (since it starts from rest again)
Acceleration (a) = 4.00 m/s^2 (remains constant throughout)
Time duration (t) = 2.0 s (total time elapsed)
v2 = 0 + (4.00 m/s^2)(2.0 s)
v2 = 0 + 8.00 m/s
v2 = 8.00 m/s
At the end of the second second, the boulder is moving at a speed of 8.00 m/s.
To find the distance traveled during the second second, we need to determine the acceleration of the boulder. We are given that the boulder has a constant acceleration.
Using the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the boulder starts from rest, the initial velocity is 0, and the equation simplifies to:
distance = (1/2) * acceleration * time^2
Given that the distance traveled during the first second is 2.00 m, and the time is 1 second, we can substitute these values into the equation:
2.00 m = (1/2) * acceleration * (1 s)^2
Simplifying the equation, we find:
acceleration = 4.00 m/s^2
To find the distance traveled during the second second, we can use the same equation, but this time with a time of 2 seconds:
distance = (1/2) * acceleration * (2 s)^2
Plugging in the values, we get:
distance = (1/2) * 4.00 m/s^2 * (2 s)^2
= (1/2) * 4.00 m/s^2 * 4 s^2
= 8.00 m
Therefore, the boulder travels 8.00 meters during the second second.
To find the velocity at the end of the first second, we can use the equation:
velocity = initial velocity + acceleration * time
Since the boulder starts from rest, the initial velocity is 0, and the equation simplifies to:
velocity = acceleration * time
Plugging in the values:
velocity = 4.00 m/s^2 * 1 s
= 4.00 m/s
Therefore, the boulder is moving at a velocity of 4.00 m/s at the end of the first second.
To find the velocity at the end of the second second, we can use the same equation, but this time with a time of 2 seconds:
velocity = acceleration * time
Plugging in the values:
velocity = 4.00 m/s^2 * 2 s
= 8.00 m/s
Therefore, the boulder is moving at a velocity of 8.00 m/s at the end of the second second.