Answer:
2.27 m/s²
Explanation:
Recall that one of the equations of motion can be expressed as
v² = u² + 2as
where
v = final velocity = takeoff speed = 33m/s
u = initial velocity = rest = 0 m/s
a = acceleration ( we are asked to find this)
s = distance traveled = length of takeoff run = 240m
substituting the known values into the equation above,
v² = u² + 2as
33² = 0² + 2a(240)
1089= 480a
a = 1089/480
a = 2.27 m/s²
ANSWER
2.27 m/s^2
EXPLANATION
We can use the general relationship from kinematics:
vf^2 = vi^2 + 2as
where:
s = distance;
a = acceleration;
vi = 0m/s is the initial velocity;
vf=33ms is the final velocity.
So:
33^2 = 0^2 + 2 ⋅ a ⋅ 240
a= 1089/480
= 2.27m/s^2
