7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
driver of the car applies the brakes after a reaction time of 0.5 s and the car slows down at a rate of
10 ms. What happens?

Respuesta :

Answer: Car collide with man

Explanation:

Given

Speed of car is [tex]u=30\ m/s[/tex]

Distance of the man from the car is [tex]s=55\ m[/tex]

Reaction time [tex]t_r=0.5\ s[/tex]

Rate of deceleration [tex]a_d=-10\ m/s^2[/tex]

Distance traveled in the reaction time [tex]d_o=30\times 0.5=15\ m[/tex]

Net effective distance to cover [tex]d=55-15=40\ m[/tex]

Distance required to stop the car

[tex]\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m[/tex]

Require distance is more than that of net effective distance. Hence, car collides with the man.