we crawled a long through the city traffic at 6m/s for 5m'. than we left the city & we gradually acceleration to 24m/s2 in 20s
What is your question?
V = Vo + a*T = 6 + 24*20 = 486 m/s.
d = Vo*T + 0.5a*T^2 = 6*300 + 0.5*24*20^2 = __meters.
To find the final velocity and distance traveled outside the city, we can use the equations of motion. The equation of motion for finding final velocity is:
v = u + at
where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
In this case, the initial velocity (u) is 6 m/s, the acceleration (a) is 24 m/s^2, and the time (t) is 20 seconds. To find the final velocity (v), we can substitute these values into the equation:
v = 6 m/s + (24 m/s^2 * 20 s)
v = 6 m/s + 480 m/s
v = 486 m/s
Therefore, the final velocity after accelerating outside the city is 486 m/s.
To find the distance traveled during this acceleration, we can use the equation:
s = ut + (1/2)at^2
where:
s is the distance traveled
u is the initial velocity
a is the acceleration
t is the time
In this case, the initial velocity (u) is 6 m/s, the acceleration (a) is 24 m/s^2, and the time (t) is 20 seconds. Substituting these values into the equation:
s = (6 m/s * 20 s) + (1/2)(24 m/s^2)(20 s)^2
s = 120 m + 1/2 * 24 m/s^2 * 400 s^2
s = 120 m + 4,800 m
s = 4,920 m
Therefore, the distance traveled outside the city during this acceleration is 4,920 meters.