Answer:
a. 5.22 meters
b. 2.9 seconds
c. No, Melvin does not hit the deer
Explanation:
The parameters with which Melvin is travelling are as follows;
The speed of Melvin's motion, u = 29 m/s
The distance from Melvin at which the deer jumps into the path = 50 m
a. Distance, d = Velocity, u × Time, t
The time it takes Melvin to react = 0.18 seconds
The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"
∴ d₁ = 29 m/s × 0.18 s = 5.22 m
The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m
b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²
Therefore, we have;
The time it takes "t₂" it takes for him to come to a complete stop given as follows;
y = u + a × t₂
Where;
v = The final velocity after Melvin comes to a complete stop = 0 m/s
By substituting the known values, we have;
0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂
∴ 29 m/s = 10 m/s² × t₂
t₂ = (29 m/s)/(10 m/s²) = 2.9 s
The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s
c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;
v² = u² + 2·a·d₂
Therefore, we have;
0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂
∴ (29 m/s)² = 2 × 10 m/s × d₂
d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m
The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m
Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.
