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An object moving 225 m/s experiences an acceleration of 62 m/s² for 26 seconds. What is the distance moved in that time?

(a) 18,200 m
(b) 19,000 m
(c) 19,800 m
(d) 20,600 m

1 Answer

5 votes

Final answer:

The distance moved by an object with an initial velocity of 225 m/s, undergoing an acceleration of 62 m/s² for 26 seconds, is 26812 meters, which suggests that there may be an error in the question or options as this answer is not listed among the choices.

Step-by-step explanation:

To calculate the distance moved by an object experiencing an acceleration, we can use the formula for motion under constant acceleration:

s = ut + (½)at2

where:

  • s is the displacement
  • u is the initial velocity
  • a is the acceleration.
  • t is the time

Given that the initial velocity (u) is 225 m/s, the acceleration (a) is 62 m/s2, and time (t) is 26 seconds, we can calculate the displacement as follows:

s = (225 m/s)(26 s) + (½)(62 m/s2)(26 s)2

s = 5850 m + (½)(62)(676)

s = 5850 m + 20962 m

s = 26812 m

Therefore, the distance moved by the object is 26812 meters, which is not listed in the given options, implying a potential error in the question or options provided.

User Mohammed Hasan
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