Answer:
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
Answer:
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
