Respuesta :
Answer:
a) [tex]\Delta s=5\ m[/tex] is the distance between deer and the vehicle
b) [tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.
Explanation:
Given:
- initial speed of driving, [tex]u=20\ m.s^{-1}[/tex]
- distance of deer from the vehicle, [tex]x=35\ m[/tex]
- reaction time taken to step onto the brakes, [tex]t'=0.5\ s[/tex]
- maximum deceleration of the car, [tex]a_m=-10\ m.s^{-2}[/tex]
a)
Now the distance travelled after application of the brakes till the vehicle stops:
[tex]v^2=u^2+2a_m.s[/tex]
(assuming that the brakes are applied with maximum acceleration)
where:
[tex]s=[/tex] displacement of the vehicle after braking till it stops
[tex]v=[/tex] final velocity of the vehicle = 0 (stops)
putting the values:
[tex]0^2=20^2-2\times 10\times s[/tex]
[tex]s=20\ m[/tex]
Now before the application of the brakes 0.5 second is taken to react and the vehicle travels during this time as well.
So, distance covered before applying the brakes:
[tex]s'=u.t'[/tex]
[tex]s'=20\times 0.5[/tex]
[tex]s'=10\ m[/tex]
The distance between the deer and the vehicle:
[tex]\Delta s=x-(s+s')[/tex]
[tex]\Delta s=35-(20+10)[/tex]
[tex]\Delta s=5\ m[/tex]
b)
The maximum speed the driver can have with the vehicle and still not hit the deer is given as:
[tex]v^2=u'^2+2. a_m.(x-s')[/tex]
because s' is the distance covered before braking during the reaction time.
[tex]0^2=u'^2-2\times 10\times (35-10)[/tex]
[tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.
