An object starts from rest with an acceleration of 2.0 m/s² in the +x-direction for 3.0 seconds. It then reduces its acceleration to 1.0 m/s² for 5.0 additional seconds. The total distance covered is about A) 52 m. B) 38 m. C) 11 m. D) 9.0 m. E) 7.6 m.

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Quizplus · Accepted Answer

The total distance covered can be calculated by finding the distance covered in each phase of the motion and then summing these distances. For the first phase (first 3.0 seconds), the object starts from rest and accelerates at 2.0 m/s^2. The distance covered in this phase can be calculated using the formula $d_1 = \frac{1}{2}at^2$, where $a = 2.0 \, \text{m/s}^2$ and $t = 3.0 \, \text{s}$, giving $d_1 = \frac{1}{2} \times 2.0 \times (3.0)^2 = 9.0 \, \text{m}$. For the second phase (next 5.0 seconds), the acceleration is 1.0 m/s^2. Using the same formula, $d_2 = \frac{1}{2} \times 1.0 \times (5.0)^2 = 12.5 \, \text{m}$. However, the initial velocity for the second phase is not zero; it's the final velocity from the first phase, which is $v = at = 2.0 \times 3.0 = 6.0 \, \text{m/s}$. The distance covered due to this initial velocity over 5.0 seconds is $v \times t = 6.0 \times 5.0 = 30.0 \, \text{m}$. Thus, the total distance for the second phase is $d_2 + 30.0 = 12.5 + 30.0 = 42.5 \, \text{m}$. Summing the distances from both phases, $d_1 + d_2 = 9.0 + 42.5 = 51.5 \, \text{m}$, which is approximately 52 m.

An Object Starts from Rest with an Acceleration of 2