A particle accelerates from rest at a uniform rate of 3 m/s^2 for a distance of 200m.
-how fast s the particle going at that time?
-how long did it take for the particle to reach that velocity?
d = (1/2)(3)t^2
so
t = sqrt [ (2/3)(200)] = sqrt (400/3)
=20/sqrt3
= 11.5 seconds
v = a t = 3 (11.5) = 34.6 m/s
To find the answers to these questions, we can use the equations of motion. The first equation relates the final velocity (v), initial velocity (u), acceleration (a), and displacement (s). It is given by:
v^2 = u^2 + 2as
In this case, the initial velocity (u) is 0 m/s, the acceleration (a) is 3 m/s^2, and the displacement (s) is 200 m. We can use this equation to find the final velocity (v).
1. How fast is the particle going at that time?
Using the equation v^2 = u^2 + 2as, we can substitute the known values:
v^2 = 0^2 + 2 * 3 * 200
v^2 = 0 + 1200
v^2 = 1200
v = √1200 ≈ 34.64 m/s
So, the particle is going approximately 34.64 m/s at that time.
2. How long did it take for the particle to reach that velocity?
To find the time (t) it took for the particle to reach that velocity, we can use the second equation of motion:
v = u + at
Again, the initial velocity (u) is 0 m/s, the acceleration (a) is 3 m/s^2, and the final velocity (v) is 34.64 m/s. Substituting these values into the equation, we get:
34.64 = 0 + 3t
3t = 34.64
t = 34.64 / 3 ≈ 11.55 seconds
So, it took approximately 11.55 seconds for the particle to reach that velocity.