A speeder traveling at 50 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 1.64 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

61s

Explanation:

the speedr travels at constant speed therefore it is the following equation

X=VT

the police travel in a uniformly accelerated movement, therefore the following equation is valid considering the initial speed is zero

X = (1/2) AT^2

as the displacement of the speeder and the policeman is the same we could match the previous equations

VT = (1/2) AT^2

Solving for T

T=2V/A

T=2 (50) / 1.64=61s

t = 61s

Explanation:

Analysis

When the policeman catches the speeder they will both have traveled the same distance d and the same time.

d₁ = d₂ = d

d₁: displacement of policeman

d₂: displacement of speede r

t₂ = t₁ = t

t₂: time of the speeder

t₁: time of the Policeman

Policeman kinematics

The policeman moves with uniformly accelerated movement:

d = v₀₁ t + (1/2) (a) t² (Formula 1)

d: displacement in meters (m)

v₀₁: initial speed = 0

a: acceleration = 1.64m/s²

t: time in s

We replace data in formula (1)

d = (1/2) (1.64) * t²

d = 0.82t² equation (1)

Speeder kinematics

The speeder moves with constant speed:

d = v * t Formula (2)

d: displacement in meters (m)

v: speed in m/s = 50 m/s

t: time in s

We replace data in formula (2)

d = 50t equation (2)

Time calculation

equation (1) = equation (2)

d = 0.82t² = 50t

0.82t = 50 (dividing by t both sides of the equation)

t = 50/0.82

t = 60.9756s = 61s