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Gemiola [76]
3 years ago
5

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

ecelerates at this same rate for the second half of the distance. If the stations are 3200m apart, find the time of travel (in seconds) between the two stations.
Physics
1 answer:
Minchanka [31]3 years ago
7 0

This problems a perfect application for this acceleration formula:

         Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:     1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half:    1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by  0.65 m/s²: 

                         T² = (1600 m) / (0.65 m/s²)

                         T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

                         T =  2 √(1600/0.65) = <em>99.2 seconds</em>  (rounded)


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<h3>What's the resultant force along x- direction?</h3>
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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

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