Answer:
v = 26.7 m/s
Explanation:
Given,
speed of the car = 20 m/s
distance between the car and the deer = 49 m
time taken to press the brake = 0.50 s
maximum deceleration of the car = 10 m/s²
Now,
distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m
distance travel by the car after the break is pressed
Using equation of motion
v² = u² + 2 a s
0² = 20² - 2 x 10 x s
s = 20 m
Total distance travel by the car = 20 + 10 = 30 m
Distance between deer and car = 49-30 = 19 m.
b. Maximum speed a car could have
Distance travel by the car in reaction time = v' x 0.5
v' is the maximum speed of the car.
maximum distance car can cover = 49 - 0.5 v'
Now, Using equation of motion
v² = u² + 2 a s
0² =v'² - 2 x 10 x (49- 0.5 x v')
v'² +10 v' -980 = 0
By solving
v = 26.7 m/s
Hence, maximum speed of the car can be 26.7 m/s
or Δ