Answer:
a. [tex]4.857m/s^2[/tex]
b.36.75 m
Explanation:
We are given that
Initial speed, u=2m/s
Final speed, v=19 m/s
Time, t=3.5 s
a.
We know that
Acceleration, a=[tex]\frac{v-u}{t}[/tex]
Using the formula
[tex]a=\frac{19-2}{3.5}[/tex]
[tex]a=\frac{17}{3.5}[/tex]
[tex]a=4.857m/s^2[/tex]
Hence, the acceleration of the rider, [tex]a=4.857m/s^2[/tex]
b.
We know that
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex](19)^2-2^2=2\times 4.857s[/tex]
[tex]361-4=9.714s[/tex]
[tex]357=9.714s[/tex]
[tex]s=\frac{357}{9.714}[/tex]
[tex]s=36.75 m[/tex]
Length of hill=36.75m
