Respuesta :
Answer:
a) t = 6.46 s, b) x = 72.98 m, c) v₁ = 26.75 m / s, v₂ = 22.61 m / s
Explanation:
This is an exercise in kinematics, let's write the expressions for each person
Hanna
leaves a time t₀o = 1s after Sam's output, both with zero initial velocity and acceleration of a₁ = 4.90 m / s²
x₁ = 0 + ½ a₁ (t-t₀) ²
v₁ = 0 + a₁ (t-t₀)
Sam
with an acceleration of a₂ = 3.50 m / s² and with an initial velocity of zero
x₂ = 0+ ½ a₂ t²
v₂ = 0 + a₂ t
a) at the point where the position of the two is found is the same
x₁ = x₂
½ a₁ (t-t₀) ² = ½ a₂ t²
let's solve
t-t₀ = [tex]\sqrt{\frac{a_2}{a_1} }[/tex] t
t (1 - [tex]\sqrt{ \frac{a_2}{a_1} }[/tex]) = t₀
t = [tex]\frac{t_o}{ 1-\sqrt{ \frac{a_2}{a_1} } }[/tex]
let's calculate
t = [tex]\frac{ 1}{1- \sqrt{\frac{3.50}{4.90} } }[/tex]
t = [tex]\frac{1}{1- 0.845}[/tex] 1 / 1- 0.845
t = 6.46 s
b) the distance traveled is
x = ½ a₂ t²
x = ½ 3.5 6.46²
x = 72.98 m
c) Hanna's speed
v₁ = 4.9 (6.46 -1)
v₁ = 26.75 m / s
sam's speed
v₂ = a2 t
v₂ = 3.50 6.46²
v₂ = 22.61 m / s
