Answer:
A) The car will hit the barrier
b) 19.82 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Distance = Speed × Time
⇒Distance = (75/3.6)×0.6
⇒Distance = 12.5 m
Distance traveled by the car during the reaction time is 12.5 m
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-\frac{75}{3.6}^2}{2\times -8.5}\\\Rightarrow s=25.53\ m[/tex]
Total distance traveled is 12.5+25.53 = 38.03 m
So, the car will hit the barrier
Distance = Speed × Time
⇒d = u0.6
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-u^2}{2\times -8.5}\\\Rightarrow s=\frac{u^2}{17}[/tex]
d + s = 35
[tex]\\\Rightarrow u0.6+\frac{u^2}{17}=35\\\Rightarrow \frac{u^2}{17}+0.6u-35=0[/tex]
[tex]10u^2+102u-5950=0\\\Rightarrow u=\frac{-51+\sqrt{62101}}{10},\:u=-\frac{51+\sqrt{62101}}{10}\\\Rightarrow u=19.82, -30.02[/tex]
Hence, the maximum velocity by which the car could be moving and not hit the barrier 35 meters ahead is 19.82 m/s.