A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
Question 1 Report
A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
Answer Details
We can use the equation of motion:
v^2 = u^2 + 2as
where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.
In this problem, the body starts from rest, so its initial velocity (u) is 0. The acceleration (a) is given as 2\(ms^{-2}\), and the displacement (s) is 9m.
Plugging in these values, we get:
v^2 = 0^2 + 2(2\(ms^{-2}\))(9m)
v^2 = 36\(ms^{-2}\)m^2
Taking the square root of both sides, we get:
v = sqrt(36\(ms^{-2}\)m^2)
v = 6\(ms^{-1}\)
Therefore, the velocity of the body after traveling 9m is 6\(ms^{-1}\). The answer is.
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