An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance does the object traverse while accelerating?A 3 mB 30 mC 90 mD 600 m
Question
An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance does the object traverse while accelerating?A 3 mB 30 mC 90 mD 600 m
Solution
To solve this problem, we can use the equation of motion: v^2 = u^2 + 2as, where:
v = final velocity u = initial velocity a = acceleration s = distance
Given in the problem: v = 30 m/s (final velocity) u = 0 m/s (initial velocity, since the object is initially at rest) a = 5 m/s^2 (acceleration)
We need to find s (distance).
Substituting the given values into the equation:
(30 m/s)^2 = (0 m/s)^2 + 2 * 5 m/s^2 * s 900 m^2/s^2 = 0 + 10 m/s^2 * s 900 m^2/s^2 = 10 m/s^2 * s
Solving for s, we get:
s = 900 m^2/s^2 / 10 m/s^2 = 90 m
So, the object traverses a distance of 90 meters while accelerating. The correct answer is C: 90 m.
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