What is the braking distance of a car traveling along a straight road at a speed of 20m / s when it brakes at an acceleration of 5m / S2?

What is the braking distance of a car traveling along a straight road at a speed of 20m / s when it brakes at an acceleration of 5m / S2?

According to 2aX = V2 − V02: x = 0 − 400 − 5 = 80m A: the braking distance is 80m

A car runs normally on a straight road at a speed of 15 m / s. when it finds a red light, it brakes at an acceleration of 1.5 m / S ^ 2. What is the displacement of the car in 15 s after braking?
Why is V0 not 15m / s but 0?
What do you mean by V0 ^ 2? The formula is not X10 = v0t + 0.5at ^ 2? Maybe I'm not a science student
Can I do this? S = V (average velocity) t = (VO + VT) t / 2 = 75?
I don't understand why V ^ 2-v0 ^ 2 = 2aX, when the final speed is zero, is v ^ 2 = 2aX? Instead of - V ^ 2 = 2aX, he doesn't add brackets, and the minus sign is out. Why is the minus sign gone?

Why is V0 not 15m / s but 0
You're not right! The initial speed of this question is VO!
v0=15
a=1.5
Braking process time t = V0 / a = 15 / 1.5 = 10s
x15=x10=vo^2/(2a)=15^2/(2*1.5)=75m
[question: what do you mean by V0 ^ 2? The formula is not X10 = v0t + 0.5at ^ 2
Can I do this? S = V (average velocity) t = (VO + VT) t / 2 = 75]
VO ^ 2 is the square of the initial velocity. V ^ 2-V 0 ^ 2 = 2 ax, V 0 ^ 2 = 2 ax when the final velocity is zero
You can do that, too
[Q2: I still don't understand that when v ^ 2-v0 ^ 2 = 2aX and the final speed is zero, why is it V0 ^ 2 = 2aX instead of - V0 ^ 2 = 2aX? He doesn't add brackets and the minus sign is out. Why is the minus sign gone
In order to simplify the calculation, the uniform deceleration motion with zero final velocity can be regarded as the uniform acceleration motion with zero initial velocity

2. The original speed of a car is 10 m / s, and then it is driven at a constant acceleration of 0.5 m / s and 178; to find the speed of the car when the acceleration is 20 s

After accelerating for 20 seconds, the speed is v = VO + at = 10 + 0.5 * 20 = 20m / s