A train starts from rest and accelerates uniformly until it has traveled 2.1 km and acquired a forward velocity of 24 m/s. The train then moves at a constant velocity of 24 m/s. For 400 s.The train then slows down uniformly at 0.065m/s². Until it is brought to a halt.The distance traveled by the train while slowing down is closest to A)4.4 km. B)4.2 km. C)4.0 km. D)3.8 km. E)3.6 km.

Question

Quizplus · Accepted Answer

The distance traveled while slowing down can be calculated using the formula $ v^2 = u^2 + 2as $, where $ v = 0 $ m/s (final velocity), $ u = 24 $ m/s (initial velocity), and $ a = -0.065 $ m/s² (deceleration). Solving for $ s $, we get $ s = \frac{u^2}{2|a|} = \frac{24^2}{2 \times 0.065} = 4.4 $ km.

A Train Starts from Rest and Accelerates Uniformly Until It