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Olenka [21]
3 years ago
10

In 1993 Ileana Salvador of Italy walked 3.0km in under 12min. Suppose that during her walk Salvador is observed to steadily incr

ease her speed from 4.20m/s to 5.00m/s in 25.0s. What is that distance traveled by Salvador during that time interval?
Physics
1 answer:
padilas [110]3 years ago
5 0

The distance covered is 115 m

Explanation:

The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:

s=(\frac{u+v}{2})t

where

s is the distance covered

u is the initiaal velocity

v is the final velocity

t is the time elapsed

In this problem, we have:

u = 4.20 m/s

v = 5.00 m/s

t = 25.0 s

Therefore, we can re-arrange the equation to find the distance covered:

s=(\frac{4.20+5.00}{2})(25.0)=115 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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