contestada

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m>s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m>s2. How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

Respuesta :

carlosego
First we write the kinematics equations:
 a
 vf = a * t + vo
 rf = a * (t ^ 2/2) + vo * t + ro
 For the bus we have:
 a = 0 (constant speed)
 vf = vo = 5
 rf = 5 * t + 12
 For the person we have
 a = 0.960
 vf = 0.960 * t
 rf = 0.960 * (t ^ 2/2)
 By the time the person reaches the back of the bus we have:
 0.960 * (t ^ 2/2) = 5 * t + 12
 0.48t ^ 2-5t-12 = 0
 Solving for t> 0
 t = 12.43 s
 Thus,
 rf = 0.960 * ((12.43) ^ 2/2) = 74.16m (the person would have to run this distance to reach the bus)
 vf = 0.960 * (12.43) = 11.9328 m / s (the person would have to go at this speed to reach the bus)
 No person is capable of running that distance at that speed.
shubhamchouhanvrVT

(a)The distance the student have to cover to reach the bus will be 74.16m

(b) The speed of the student to reach the bus will be 11.9 m/s

What is acceleration?

Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction

First we write the kinematics equations:

[tex]v_f = a t + v_o[/tex]

[tex]r_f = a (\dfrac{t ^ 2}{2}) + v_o t + r_o[/tex]

For the bus we have:

a = 0 (constant speed)

vf = vo = 5

rf = 5 * t + 12

For the person we have

a = 0.960

vf = 0.960 * t

rf = 0.960 * (t ^ 2/2)

By the time the person reaches the back of the bus we have:

[tex]0.960 \times (\dfrac{t ^ 2}{2}) = 5 t + 12[/tex]

[tex]0.48t ^ 2-5t-12 = 0[/tex]

Solving for t> 0

t = 12.43 s

Thus,

[tex]r_f = 0.960 \times ((\dfrac{12.43) ^ 2}{2}) = 74.16m[/tex]

(the person would have to run this distance to reach the bus)

[tex]v_f = 0.960 \times (12.43) = 11.9328 \frac{m} { s}[/tex]

(the person would have to go at this speed to reach the bus)

No person is capable of running that distance at that speed.

To know more about acceleration follow

https://brainly.com/question/25749514

Otras preguntas