Answer:
v = 26. 88 m/s +23 m/s
Explanation:
u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m
[tex]\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}[/tex]
[tex]\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}[/tex]
Solve s'
[tex]\frac{1}{s'} = \frac{2}{1.50 m} - \frac{1}{2.0 m}[/tex]
[tex]\frac{1}{s'} = 1.833 m[/tex]
[tex]s' = - 0.545 m[/tex]
To determine the speed of the trick to the highway
[tex]\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}[/tex]
[tex]\frac{ds}{dt} = 26.88 m/s[/tex]
Now to determine the velocity highway is going to be
v = ds/dt + u
v = 26. 88 m/s +23 m/s
