QuestionJuly 26, 2025

An object starts from rest and travels in a straight line with an acceleration of 5.0 m/s^2 for 4.0 seconds. I then reduces its acceleration to 2.7m/s^2 for an additional 8.5 seconds. The magnitude of the velocity at the end of the 8.5 second interval is Your Answer: square square Answer units

An object starts from rest and travels in a straight line with an acceleration of 5.0 m/s^2 for 4.0 seconds. I then reduces its acceleration to 2.7m/s^2 for an additional 8.5 seconds. The magnitude of the velocity at the end of the 8.5 second interval is Your Answer: square square Answer units
An object starts from rest and travels in a straight line with an acceleration of 5.0 m/s^2 for 4.0 seconds. I then reduces its acceleration to 2.7m/s^2 for an additional 8.5 seconds. The magnitude of the velocity at the end of the 8.5 second interval is Your Answer: square square Answer units

Solution
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Answer

42.95 m/s Explanation 1. Calculate velocity after first interval Use **v = u + at**. Initial velocity u = 0, acceleration a = 5.0 \, m/s^2, time t = 4.0 \, s. So, v = 0 + (5.0)(4.0) = 20.0 \, m/s. 2. Calculate velocity after second interval Use **v = u + at** again. Now, initial velocity u = 20.0 \, m/s, acceleration a = 2.7 \, m/s^2, time t = 8.5 \, s. So, v = 20.0 + (2.7)(8.5) = 42.95 \, m/s.

Explanation

1. Calculate velocity after first interval<br /> Use **$v = u + at$**. Initial velocity $u = 0$, acceleration $a = 5.0 \, m/s^2$, time $t = 4.0 \, s$. So, $v = 0 + (5.0)(4.0) = 20.0 \, m/s$.<br /><br />2. Calculate velocity after second interval<br /> Use **$v = u + at$** again. Now, initial velocity $u = 20.0 \, m/s$, acceleration $a = 2.7 \, m/s^2$, time $t = 8.5 \, s$. So, $v = 20.0 + (2.7)(8.5) = 42.95 \, m/s$.
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