Question

Hannah tests her new sports car by racing with Sam, an experienced racer. Both start from rest, but Hannah leaves the starting line 1.00 s after Sam does. Sam moves with a constant acceleration of 3.50 m/s2, while Hannah maintains an acceleration of 4.90 m/s2. Find (a) the time at which Hannah overtakes Sam, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant Hannah overtakes Sam.

Answer 1

a) t = 6.46 s, b) x = 72.98 m, c) v₁ = 26.75 m / s, v₂ = 22.61 m / s

Step-by-step 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 =
`\sqrt{(a_2)/(a_1) }` t

t (1 -
`\sqrt{ (a_2)/(a_1) }`) = t₀

t =
`\frac{t_o}{ 1-\sqrt{ (a_2)/(a_1) } }`

let's calculate

t =
`\frac{ 1}{1- \sqrt{(3.50)/(4.90) } }`

t =
`(1)/(1- 0.845)` 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