Answer:
d= 1400 ft :Distance from the starting point where the car reaches the truck
Explanation:
Car Kinematics : the car has uniformly accelerated movement, so we apply the following equation for distance
d₁= v₀₁*t + (1/2)* a₁*t₁² Equation 1
d₁: car distance (ft)
v₀₁: car initial speed (ft/s)
a₁ : car acceleration (ft/s²)
t1 : car time (s)
Truck Kinematics: The truck moves with uniform motion (constant speed), so the equation for distance is:
d₂= v₂*t₂ Equation (2)
d₂: truck distance (ft)
v₂: truck speed (ft/s)
t₂ : truck time (s)
Known data
v₀₁ = 0
a₁ = 7 ft/s²
v₂ = 70 ft/s
Problem development
The time and distance for when the car catches the truck is the same, then:
d₁=d₂ and t₁=t₂= t
Equation (1 )= the Equation ( 2)
v₀₁*t₁ + (1/2)* a₁*t₁² = v₂*t₂
v₀₁*t + (1/2)* a₁*t²= v₂*t
0*t + (1/2)*7*t² = 70*t : we divide both sides of the equation by t
(1/2)*7*t = 70
t= 20 s
We replace t= 20 s in the equation (2)
d= 70*20 = 1400 ft :Distance from the starting point where the car reaches the truck
Answer:112.82 m/s
Explanation:
Given
range of arrow=68 m
as the arrow travels it acquire a vertical velocity
-------1
Range is given by
R=ut
where u=initial velocity
substitute the value of t in eqn 1
--------2
and
substitute it in 2
u=112.82 m/s