Respuesta :
Answer:
given,
speed of the car = 20 m/s
final speed of car = 0 m/s
distance between car and the deer = 38 m
reaction time, t = 0.5 s
deceleration of the car = 10 m/s².
a) distance between deer and car
distance travel in the reaction time
d₁ = v x t
d₁ = 20 x 0.5 = 10 m
distance travel after you apply brake
using equation of motion
v² = u² + 2 a s
0 = 20² - 2 x 10 x s
s = 20 m
total distance traveled by the car
D = d₁ + d₂
D = 20 + 10 = 30 m
distance between car and the deer = 38 m - 30 m
= 8 m
b) now, maximum speed car.
distance travel in reaction time
d₁ = s x t
d₁ = 0.5 V
distance left between them
d₂ = 38 - d₁
d₂ = 38 - 0.5 V
distance travel after you apply brake
using equation of motion
v² = u² + 2 a d₂
0 = (V)² - 2 x 10 x (38 - 0.5 V)
V² + 10 V - 760 = 0
now, solving the quadratic equation
[tex]x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}[/tex]
V = 23.01 , -33.01
rejecting the negative term.
hence, maximum speed of the car could be V = 23.01 m/s
