Constant acceleration: An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 12.0 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object? A) -12.0 m/s² B) 24.0 m/s² C) -4.00 m/s² D) 4.00 m/s² E) 8.00 m/s²

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Quizplus · Accepted Answer

The displacement from t = 1.00 s to t = 2.00 s is given as 12.0 m. Using the kinematic equation for displacement, $ s = ut + \frac{1}{2}at^2 $, where initial velocity $ u = 0 $ at t = 0, we can find the acceleration. The displacement from t = 0 to t = 2 s is $ s_2 = \frac{1}{2}a(2)^2 = 2a $. From t = 0 to t = 1 s, $ s_1 = \frac{1}{2}a(1)^2 = 0.5a $. The displacement from t = 1 to t = 2 s is $ s_2 - s_1 = 12 $, so $ 2a - 0.5a = 12 $, giving $ 1.5a = 12 $, thus $ a = 8 $ m/s².

Constant Acceleration: an Object Starts from Rest at Time T