Question

Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t = 0.0. Knowing that automobile A has a constant acceleration of 0.8 m/s? and automobile B has a constant deceleration of 0.4 m/s2. Automobile A will overtake B after traveling a distance SA: A B. Side view​

Answer 1

x = 240 m

Step-by-step explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

the initial position of car a is zero

x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

x = x_{ob} + v_{ob} t - ½ a_b t²

car B's starting position is 30 m

x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

v_{ob} = 23.4 km / h = 6.5 m / s

4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

0.2 t² - 2.5 t - 30 = 0

t² - 12.5 t - 150 = 0

we solve the quadratic equation

t =
`(12.5 \pm √(12.5^2 + 4 \ 150) )/(2)`

t =
`(12.5 \ \pm 27.5)/(2)`

t₁ = 20 s

t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

x = 4 t + ½ 0.8 t²

x = 4 20 + ½ 0.8 20²

x = 240 m