A motor of a certain elevator gives it a constant upward acceleration of 0.767 m/s2, the elevator starts from rest, accelerates for 2s, then continues with a constant speed.

a) What is his final speed after 2s?
b) Calculate speed after 1 ,and 3 s.?
c) How far has it risen 1,2,3,and 5 s after start?

To solve this problem, we can use the equations of motion that relate the initial velocity (V₀), final velocity (V), acceleration (a), and time (t).

a) To find the final speed after 2 seconds, we can use the formula:

V = V₀ + at

Given that the elevator starts from rest (V₀ = 0 m/s) and has a constant upward acceleration of 0.767 m/s², the calculation becomes:

V = 0 + (0.767 m/s²) * (2 s) = 1.534 m/s

Therefore, the final speed of the elevator after 2 seconds is 1.534 m/s.

b) To calculate the speed after 1 second, we can use the same formula:

V = V₀ + at

Given that the elevator starts from rest (V₀ = 0 m/s) and has a constant acceleration of 0.767 m/s², we can calculate the speed after 1 second:

V = 0 + (0.767 m/s²) * (1 s) = 0.767 m/s

Therefore, the speed of the elevator after 1 second is 0.767 m/s.

To calculate the speed after 3 seconds, we can use the formula as well:

V = V₀ + at

Given that the elevator starts from rest (V₀ = 0 m/s) and has a constant acceleration of 0.767 m/s², we can calculate the speed after 3 seconds:

V = 0 + (0.767 m/s²) * (3 s) = 2.301 m/s

Therefore, the speed of the elevator after 3 seconds is 2.301 m/s.

c) To calculate the distance the elevator has risen after a certain time, we can use the kinematic equation:

d = V₀t + (1/2)at²

where d is the distance, V₀ is the initial velocity, t is the time, and a is the acceleration.

To find the distance after 1 second:

d = (0 m/s) * (1 s) + (1/2) * (0.767 m/s²) * (1 s)² = 0.3835 m

To find the distance after 2 seconds:

d = (0 m/s) * (2 s) + (1/2) * (0.767 m/s²) * (2 s)² = 1.534 m

To find the distance after 3 seconds:

d = (0 m/s) * (3 s) + (1/2) * (0.767 m/s²) * (3 s)² = 3.4515 m

To find the distance after 5 seconds:

d = (0 m/s) * (5 s) + (1/2) * (0.767 m/s²) * (5 s)² = 9.1675 m

Therefore, the distances the elevator has risen after 1, 2, 3, and 5 seconds are 0.3835 m, 1.534 m, 3.4515 m, and 9.1675 m, respectively.

To solve this problem, we will use the following kinematic equations:

1. Final velocity (v) = Initial velocity (u) + Acceleration (a) x Time (t)
2. Displacement (s) = Initial velocity (u) x Time (t) + 0.5 x Acceleration (a) x Time^2 (t^2)

a) What is the final speed after 2 seconds?
To find the final speed after 2 seconds, we can use equation 1. Given that the initial velocity (u) is 0, the acceleration (a) is 0.767 m/s^2, and the time (t) is 2 seconds, the equation becomes:

v = 0 + 0.767 m/s^2 x 2 s = 1.534 m/s

Therefore, the final speed after 2 seconds is 1.534 m/s.

b) Calculate the speed after 1 and 3 seconds.
To find the speed after 1 second, we can use equation 1 again. Given that the initial velocity (u) is 0, the acceleration (a) is 0.767 m/s^2, and the time (t) is 1 second, the equation becomes:

v = 0 + 0.767 m/s^2 x 1 s = 0.767 m/s

Therefore, the speed after 1 second is 0.767 m/s.

To find the speed after 3 seconds, we can use equation 1 again. Given that the initial velocity (u) is 0, the acceleration (a) is 0.767 m/s^2, and the time (t) is 3 seconds, the equation becomes:

v = 0 + 0.767 m/s^2 x 3 s = 2.301 m/s

Therefore, the speed after 3 seconds is 2.301 m/s.

c) How far has it risen after 1, 2, 3, and 5 seconds?
To find the distance traveled, we can use equation 2. Given that the initial velocity (u) is 0, the acceleration (a) is 0.767 m/s^2, and the time (t) is 1, 2, 3, and 5 seconds respectively, the equation becomes:

For 1 second:
s = 0 x 1 s + 0.5 x 0.767 m/s^2 x (1 s)^2 = 0.3835 m

For 2 seconds:
s = 0 x 2 s + 0.5 x 0.767 m/s^2 x (2 s)^2 = 1.534 m

For 3 seconds:
s = 0 x 3 s + 0.5 x 0.767 m/s^2 x (3 s)^2 = 3.45 m

For 5 seconds:
s = 0 x 5 s + 0.5 x 0.767 m/s^2 x (5 s)^2 = 9.175 m

Therefore, the elevator has risen 0.3835 m after 1 second, 1.534 m after 2 seconds, 3.45 m after 3 seconds, and 9.175 m after 5 seconds.