Calculus question

Question

A cyclist traveling at 40ft/s decelerates at a constant 4ft/s^2. How many feet does she travel before coming to a complete stop?

Answer 1

Hey Jo,

 

First, define what speed and accelerations are:

 

speed or velocity = dx / dt     accelearation = dv / dt

 

dv = dx / dt     (dv) (dt) = dx    a = dv / dt

 

a will be negative since it is decelerating:

 

a = dv / dt  -->  a = (vf - vi) / dt   -->   -4 = (0 - 40) / dt   -4 = -40 / dt

 

-4 dt = (-40 / dt) dt   -->   -4dt = -40   -->   dt = 10   t = 10

 

 

v_f = v_, + at    (to find out time)

 

v_f - v_i = at   --->   dv = at

 

Integrate whole equation respect to t:

 

(Since it is integrating respect to t, v and a are constant)

 

∫ (v_f = v_i + at) dt  = ∫  v_f dt = ∫ (v_i + at) dt

 

∫ v_f dt = ∫ v_i dt + ∫ at dt

 

dv = dx / dt

 

∫ v_f dt - ∫ v_i dt = ∫ at dt     (Just to show you how the equation below is worked out)

 

x = v_i t + (1/2) at²    Do you remember this equation (formula) from Physics?

 

(to find out the distance) (after finding time from 1st equation)

 

x = 40 (10) + (1/2) (-4)(10²) = 400 - 200 = 200   --->   200ft

 

∫ v_f dt - ∫ v_i dt = ∫ at dt

 

∫ (v_f - v_i) dt = at² / 2

 

∫ (dv) dt = at² / 2

 

∫ (dx / dt) dt = at² / 2

 

∫ dx = at² / 2

 

x = at² / 2   x = 4 (10)² /2 = 4 (100) / 2 = 400 / 2 = 200

 

200 ft.

 

Good luck to you.