Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

A train with an initial velocity of 31 m/s begins accelerating at rate of 0.0705 m/s^2. If the train travels for 180.5s, how far does it travel?

Sagot :

Answer:

v = 43.72 m/s

Explanation:

Given that,

Initial velocity of the train, u = 31 m/s

Acceleration of the train, a = 0.0705 m/s²

Time for which the train travel, t = 180.5 s

We need to find the final velocity of the train. Let it is v.. Using first equation of kinematics to find it such that,

[tex]v=u+at\\\\v=31+0.0705\times 180.5\\\\v=43.72\ m/s[/tex]

So the final speed of the train is equal to 43.72m/s.

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.