Answer:
25 seconds
Explanation:
Assuming the woman is accelerating at a constant rate of [tex]a \;\;m/s^2[/tex] from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.
Let she takes t seconds to cover the distance, s=115 m.
As acceleration, [tex]a=\frac{v-u}{t}=\frac{5-4.2}{t}[/tex]
[tex]\Rightarrow at=0.8\cdots(i)[/tex]
Now, from the equation of motion
[tex]s=ut+\frac 12 at^2[/tex]
[tex]\Rightarrow s=ut+\frac 12 at(t)[/tex]
[tex]\Rightarrow 115=4.2t+\frac 12 \times 0.8 t[/tex] [ from equation (i)]
[tex]\Rightarrow 115=(4.2+0.4)t[/tex]
[tex]\Rightarrow t= 115/4.6 = 25[/tex] seconds.
Hence, she takes 25 seconds to walk the distance.
