Atom Ant is traveling with an initial velocity of 20.0 cm/s. He begins to accelerate at a rate of 8.00 cm/s2 for 5.00 s. What is his total displacement in the 5.00 second interval? What is his displacement in the last second?
d= vi*t+1/2 a t^2
vi is given, a is given, in the last second
find d for t=5 seconds, then t=4 seconds, the difference is the displacment in the last second.
To solve this problem, we can use the equations of motion. The equation we'll need is:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 20.0 cm/s
acceleration (a) = 8.00 cm/s^2
time (t) = 5.00 s
1. Total Displacement:
Using the equation of motion:
displacement = 20.0 cm/s * 5.00 s + (1/2) * 8.00 cm/s^2 * (5.00 s)^2
displacement = 100.0 cm + (1/2) * 8.00 cm/s^2 * 25.00 s^2
displacement = 100.0 cm + 100.0 cm
displacement = 200.0 cm
Therefore, Atom Ant's total displacement in the 5.00 second interval is 200.0 cm.
2. Displacement in the Last Second:
To find the displacement in the last second, we can use the same equation of motion, but substitute the time with the last second, which is 1.00 second:
displacement = 20.0 cm/s * 1.00 s + (1/2) * 8.00 cm/s^2 * (1.00 s)^2
displacement = 20.0 cm + (1/2) * 8.00 cm/s^2 * 1.00 s^2
displacement = 20.0 cm + 4.00 cm
displacement = 24.0 cm
Therefore, Atom Ant's displacement in the last second is 24.0 cm.
To solve this problem, we can use the equations of motion.
The first step is to find the final velocity using the equation:
vf = vi + a * t
where
vf is the final velocity,
vi is the initial velocity,
a is the acceleration, and
t is the time interval.
Given:
vi = 20.0 cm/s
a = 8.00 cm/s^2
t = 5.00 s
Substituting the given values into the equation:
vf = 20.0 cm/s + (8.00 cm/s^2 * 5.00 s)
vf = 20.0 cm/s + 40.0 cm/s
vf = 60.0 cm/s
Now, we can find the displacement using the equation:
Δx = vi * t + 0.5 * a * t^2
where
Δx is the displacement
Substituting the given values into the equation:
Δx = 20.0 cm/s * 5.00 s + (0.5 * 8.00 cm/s^2 * (5.00 s)^2)
Δx = 100.0 cm + (0.5 * 8.00 cm/s^2 * 25.00 s^2)
Δx = 100.0 cm + 100.0 cm
Δx = 200.0 cm
Therefore, the total displacement of Atom Ant in the 5.00 second interval is 200.0 cm.
To find the displacement in the last second, we subtract the initial displacement from the total displacement. Since Atom Ant starts with an initial velocity of 20.0 cm/s, his initial displacement is given by:
initial displacement = vi * t
initial displacement = 20.0 cm/s * 5.00 s
initial displacement = 100.0 cm
Therefore, the displacement in the last second is:
displacement in the last second = total displacement - initial displacement
displacement in the last second = 200.0 cm - 100.0 cm
displacement in the last second = 100.0 cm
So, Atom Ant's displacement in the last second is 100.0 cm.