Question

An object which is initially at rest accelerates at a rate of 6.0 m/s. Its final position is 44 m from its initial position at the end of that acceleration. For how much time did it accelerate?

a) 7.3 seconds
b) 11.0 seconds
c) 22.0 seconds
d) 66.0 seconds

Answer 1

To calculate the time an object at rest accelerates to travel 44 m with an acceleration of 6.0 m/s², the kinematic equation s = ut + ½at² is used. After plugging in the values, the timeframe is calculated to be approximately 3.83 seconds, which doesn't match any of the provided options.

Step-by-step explanation:

To determine the time an object with constant acceleration takes to travel a certain distance from rest, one can use the kinematic equation:

s = ut + ½at2

where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. In this problem, the object starts from rest, which means u = 0 m/s, a = 6.0 m/s2, and s = 44 m.

Substituting the given values:

44 m = (0 m/s)(t) + ½(6.0 m/s2)(t2)

Therefore, we get:

44 m = 3.0 m/s2(t2)

t2 = 44 m / 3.0 m/s2 = 14.67 s2

t = √14.67 s2

t ≈ 3.83 s

However, none of the given options (7.3 seconds, 11.0 seconds, 22.0 seconds, 66.0 seconds) match the calculated time of approximately 3.83 seconds, suggesting either a misinterpretation of the question or potential error in the provided options.