A body accelerates uniformly from rest at 2m/s².calculate its velocity after travelling 9m
v = at
plug in your numbers
think about it. every second, the velocity increases by 2 m/s.
vi = 0 m/s,
Δd = 9 m,
a = 2 m/s²,
vf = ?
vf² = vi² + 2aΔd
vf = sqrt(vi² + 2aΔd)
vf = sqrt[0² + 2(2)(9)]
vf = sqrt[36]
∴vf = 6 m/s
To calculate the velocity of a body accelerating uniformly, we can use the equation:
v² = u² + 2as,
where:
v = final velocity of the body,
u = initial velocity of the body (in this case, the body starts from rest, so u = 0),
a = acceleration of the body, and
s = displacement of the body.
In this case, the acceleration is given as 2 m/s², and the displacement is given as 9 m. Since the body starts from rest, the initial velocity (u) is 0.
Plugging these values into the equation,
v² = 0² + 2(2)(9)
v² = 36
Taking the square root of both sides,
v = √36
v = 6
Therefore, the velocity of the body after traveling 9 m is 6 m/s.
V^2 = Vo^2 + 2a*d = 0 + 4*9 = 36,
V = ___m/s.