An object is at rest at #(7 ,4 ,2 )# and constantly accelerates at a rate of #4/3 m/s# as it moves to point B. If point B is at #(3 ,1 ,9 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Apr 27, 2016

Answer:

#t_F ~~ 3.592" "# to 3 decimal places

Explanation:

#color(red)("Note that I have shown how to deal with units")#

#color(green)("Tip:- leave the converting to decimal until the end")#

#color(blue)("Determine the distance of travel")#

Let distance be #d#
Let initial time be #t_0# seconds
Let time at any instant #i# be #t_i# seconds
Let final time be #t_F# seconds

Given: acceleration #-> 4/3color(white)(.)color(red)( m/s^2)#

Let point 1 be #P_1->(x_1,y_1,z_1)->(7,4,2)#
Let point 2 be #P_2->(x_2,y_2,z_2)->(3,1,9)#

Using Pythagoras

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

#d=sqrt((3-7)^2+(1-4)^2+(9-2)^2)#

#d=sqrt(16+9+49)#

#d=sqrt(74)" "# metres#" "#(74 is a product of 2 primes)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine time of travel")#

Velocity at #t_0=0color(white)(.) m/s#

Velocity at #t_i = 4/3 xx t_i color(green)(" units"-> m/s^sxxs color(blue)(->mxxs/s^2=m/s))#

Mean velocity #1/2xx4/3xxt_F color(white)(.)color(blue)(m/s)#

Thus distance is mean velocity times time giving

#" "d=1/2xx4/3xxt_Fxxt_F #

#" "d= 4/6(t_F)^2 color(green)(" units"-> m/sxxs ->color(blue)(mxxs/s=m#

Multiply both sides by #6/4#

#" "d6/4=4/6xx6/4(t_F)^2#

But #4/6xx6/4=1#

#" "=>d6/4=(t_F)^2#

Thus

#" "t_F=sqrt(6/4)xxsqrt(sqrt(74))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider #sqrt(6/4) -> sqrt((6)/(2^2))=1/2sqrt(6)#

Consider #sqrt(sqrt(74)) -> root(4)(74)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Putting it all together")#

#" "t_F=1/2sqrt(6)xxroot(4)(74)#

By calculator: #t_F ~~ 3.592" "# to 3 decimal places