Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship

(1)
where
a is the deleceration
S is the distance covered after a time t

is the velocity at time t

is the initial speed of the car
The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and

(the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve the equation (1) to find a:

And the negative sign means it is a deceleration.
As the speed is a scalar quantity as it has the only magnitude in it. Therefore, the average speed is also stated as a scalar quantity.
Hence, the correct answer is (B)
Answer:
6.77 m/s
Explanation:
First, in the x direction:
Given:
Δx = 3.17 m
v₀ = v cos 30.8° = 0.859 v
a = 0 m/s²
Δx = v₀ t + ½ at²
(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²
3.17 = 0.859 v t
3.69 = v t
Next, in the y direction:
Given:
Δy = 0.432 m
v₀ = v sin 30.8° = 0.512 v
a = -9.81 m/s²
Δy = v₀ t + ½ at²
(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²
0.432 = 0.512 v t − 4.905 t²
Two equations, two variables. Solve for t in the first equation and substitute into the second equation:
t = 3.69 / v
0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²
0.432 = 1.89 − 66.8 / v²
66.8 / v² = 1.458
v² = 45.8
v = 6.77
Proton I think if I remember right