An object is at rest at #(7 ,6 ,4 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(9 ,5 ,7 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Shwetank Mauria
Jun 23, 2016

It will take #2.45# seconds.

Explanation:

The distance between two points #(x_1,y_1,z_1)# and #(x_2,y_2,z_2)# is given by

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

Hence distance between #(7,6,4)# and #(9,5,7)# is

#sqrt((9-7)^2+(5-6)^2+(7-4)^2)#

= #sqrt(2^2+1^2+3^2)=sqrt(4+1+9)=sqrt14#

(As distance covered is given by #S=ut+1/2at^2#, where #u# is initial velocity, #a# is accelaration and #t# is time taken. If body is at rest #S=1/2at^2# and hence #t=sqrt((2S)/a)#

As the coordinates are in meters, the time taken at an acceleration of #5/4# #m/sec^2# will be given by

#t=sqrt((2sqrt14)/(5/4))=sqrt((2sqrt14)xx(4/5))=sqrt((8sqrt14)/5))#

= #sqrt(8xx3.74166/5)=sqrt5.986656=2.45#

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