Question

A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. How far will it travel in 20 s? What will its final velocity be

Answer 1

Answer: A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.

Explanation: To find the answer, we need to know about uniformly accelerated motion.

How to solve the problem?

• Given that,

                     $u=16m/s\\a=2 m/s^2\\t=20s$

• We have to find the distance travelled by the train.

• As we have,

                      $S=ut+\frac{1}{2}at^2$

• Substituting values, we get,

                      $S=(16*20)+\frac{2*20^2}{2} =720 m.$

• We have the equation for final velocity as,

                    $v^2=u^2+2aS\\thus,\\v=\sqrt{u^2+2aS} =\sqrt{16^2+(2*2*720)} =56 m/s.$

Thus, we can conclude that, a train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.

Learn more about the uniformly accelerated motion here:

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Answer 2

A train experiencing constant acceleration of 2 m/s^2 is moving at an initial speed of 16 m/s. It will go 720 meters in 20 seconds, with a final velocity of 56 m/s.

Understanding uniformly accelerated motion is necessary in order to determine the solution.

How can the issue be resolved?

• We need to determine how far the train has traveled. We have,

                         $S=ut+\frac{1}{2}at^2 \\S=720m\\where,\\u=16m/s, a=2m/s^2,t=20s$

• The formula for final velocity is as follows:

                          $v^2-u^2=2aS\\v=\sqrt{u^2+2aS} \\v=56m/s$

Thus, we may say that a train moving at 16 m/s initially experiences constant acceleration of 2 m/2. It will go 720 meters in 20 seconds, with a final velocity of 56 meters per second.

Learn more about uniformly accelerated motion here:

brainly.com/question/28105762

#SPJ1