Question

A motorist is driving at 15 m/s when she sees that a traffic light 315m ahead has just turned red. She knows that this light stay red for 25 s, and she wants to be 20 m from the ligt when it turns green again. Taht way, she will still be able to stop if the light stays red longer than expected. She applies the brake gradually such that her acceleration is ax(t)= c + bt, where c and b are constant. Assume she starts with a constant speed at the origin.

Find the values of c a b and any other unknown constants in order to answer the following questions.

1. Given the motorist's acceleration as a function of time, what are her position and velocity fucntions? - Do not use numbers for any constant here. Only derive the position and velocity functions.

2. What is her speed as she reaches the light?

Answer 1

1) x = x₀ + vot - ½ c t² - 1/6 bt³, v = v₀ - ct - ½ b t²

2) v₁ = 5.25 m/s, v₂ = -8 m/s

Step-by-step explanation:

1) For this exercise, the relationship of the body is not constant, so you must use the definition of speed and position to find them.

acceleration is

a = c + bt

a) the relationship between velocity and acceleration

a =
`(dv)/(dt)`

dv = -a dt

The negative sign is because the acceleration is contrary to the speed to stop the vehicle.

we integrate

∫ dv = - ∫ a dt

∫ dv = -∫ (c + bt) dt

v = -c t - ½ b t²

This must be valued from the lower limit, the velocity is vo, up to the upper limit, the velocity is v for time t

v - v₀ = -c (t-0) - ½ b (t²-0)

v = v₀ - ct - ½ b t²

b) the velocity of the body is

v =
`(dx)/(dt)`

dx = v dt

we replace and integrate

∫ dx = ∫ (v₀ - c t - ½ bt²) dt

x-x₀ = v₀ t - ½ c t² - ½ b ⅓ t³

Evaluations from the lower limit the body is at x₀ for t = 0 and the upper limit the body is x = x for t = t

x - x₀ = v₀ (t-0) - ½ c (t²-0) +
`(1)/(6)` (t³ -0)

x = x₀ + vot - ½ c t² - 1/6 bt³

2) The speed when you reach the traffic light

Let's write the data that indicates, the initial velocity is vo = 15 m / s, the initial position is xo = 315m, let's use the initial values ​​to find the constants.

t = 25 s x = 20

we substitute

20 = 315 + 15 25 - ½ c 25² - 1/6 b 25³

0 = 295 + 375 - 312.5 c - 2604.16 b

670 = 312.5 c + 2604.16 b

we simplify

2.144 = c + 8.33 b

Now let's use the equation for velocity,

v = v₀ - ct - ½ b t²

v = 15 - c 25 - ½ b 25²

v = 15 - 25 c - 312.5 b

let's write our two equations

2.144 = c + 8.33 b

v = 15 - 25 c - 312.5 b

Let's examine our equations, we have two equations and three unknowns (b, c, v) for which the system cannot be solved without another equation, in the statement it is not clear, but the most common condition is that if the semaphore does not change, it follows with this acceleration (constant) to a stop

a = c + b 25

from the first equation

c = 8.33 / 2.144 b

C = 3.885 b

we substitute in the other two

v = 15 - 25 (3.885 b) - 312.5 b

v = 15 - 409.6 b

final acelearation

a = 28.885 b

let's use the cinematic equation

`v_(f)^2`= v² - 2 a x

0 = v² - 2a 20

0 = v² - (28.885b) 40

v² = 1155.4 b

we write the system of equations

v = 15 - 409.6 b

v² = 1155.4 b

resolve

v²= 1155.4 (
`(15 -v )/(409.6)` )

v² = 2.8 ( 15 -v)

v² + 2.8 v - 42.3 = 0

v= [ -2.8 ±
`\sqrt {2.8^2 + 4 \ 42.3) }` ]/2 = [-2.8 ± 13.3]/2

v₁ = 5.25 m/s

v₂ = -8 m/s