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- The school zone in front of your school has a posted speed limit of 25 mi/h, which is about 11 m/s. Let's
examine the stopping of a car in several different situations.
a. The crossing guard holds up her stop sign, and the driver is paying attention well. The car moves at a
constant velocity of 11 m/s for 2.3 seconds while the driver reacts, then slows down at a constant rate
of -4.5 m/s2. What is the stopping distance for the car in this situation?​

Respuesta :

hamzaahmeds

Answer:

s = 38.7 m

Explanation:

First we calculate the distance covered during uniform motion of reaction time.

s₁ = vt

where,

s₁ = distance covered during uniform motion = ?

v = uniform speed = 11 m/s

t = time = 2.3 s

Therefore,

s₁ = (11 m/s)(2.3 s)

s₁ = 25.3 m

Now, we calculate the distance covered during decelerated motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = -4.5 m/s²

s₂ = distance covered during decelerated motion = ?

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 11 m/s

Therefore,

2(-4.5 m/s²)s₂ = (0 m/s)² - (11 m/s)²

s₂ = (-121 m²/s²)/(-9 m/s²)

s₂ = 13.4 m

the total distance will be:

s = s₁ + s₂

s = 25.3 m + 13.4 m

s = 38.7 m

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