Question

In 1993 Ileana Salvador of Italy walked 3.0km in under 12min. Suppose that during her walk Salvador is observed to steadily increase her speed from 4.20m/s to 5.00m/s in 25.0s. What is that distance traveled by Salvador during that time interval?

Answer 1

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:

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

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