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Mariana [72]
2 years ago
7

A car traveling 85 km/h slows down at a constant 0.53 m/s^2 just by "letting up on the gas." Calculate the distance it travels d

uring the first second.
Calculate the distance it travels during the fifth second.
Physics
1 answer:
snow_lady [41]2 years ago
7 0

Answer:

a) d = 23.345 m : distance the car travels in the first second

b) d = 111.425 m : distance the car travels in the fifth second

Explanation:

The distance (d) in uniformly accelerated motion is calculated as follows:

d = v₀*t +(1/2)*a*t² Formula (1)

d: distance (m)

v₀: initial speed (m/s)

a: acceleration (m/s²)

t: time (s)

Equivalences

1 km = 1000m

1 h = 3600 s

Data

v₀ = 85 km/h = 85*1000m/3600s = 23.61 m/s

a = - 0.53 m/s²

Calculation of the distance the car travels in the first second

We replace data in formula (1) at t = 1s

d = 23.61*(1)+ (1/2)*(-0.53)*(1)²

d = 23.345 m

Calculation of the distance the car travels in the fifth second

We replace data in formula (1) at t = 5s

d = 23.61*(5)+ (1/2)*(-0.53)*(5)²

d = 111.425 m

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