Answer:
11.4 m
Explanation:
We are given that
Total distance traveled by cab,d=214 m
Maximum speed of cab=v=315 m/min=[tex]\frac{315}{60}=5.25m/s[/tex]
1 min =60 s
Acceleration of cab=[tex]1.21 m/s^2[/tex]
We have to find the distance traveled by the cab while accelerating to full speed from rest.
Initial speed o f cab=u=0
[tex]v^2-u^2=2ax[/tex]
Substitute the values in the formula
[tex](5.25)^2-0=2(1.21)x[/tex]
[tex]27.5625=2.42x[/tex]
[tex]x=\frac{27.5625}{2.42}[/tex]
[tex]x=11.4 m[/tex]
Hence, the cab travel 11.4 m from rest to full speed.