A body starts from rest and is uniformly accelerated for 30 sec.The distance covered in first 10 sec is #x_1# , for next 10 sec is #x_2# and for last 10 sec is #x_3# the then the ratio of #X_1:X_2:X_3#=?

2 Answers
Apr 11, 2017

Answer:

#k=0,1,2,3,....#
#1:3:5:7:(2k+1)#

Explanation:

  • please look at the animation and calculate the colored areas.

enter image source here
#k=0,1,1,3,4...#
#1:3:5:7:(2k+1)#

Apr 12, 2017

Answer:

#x_1:x_2:x_3=1:3:5#

Explanation:

The applicable kinematic expression is
#s=ut+1/2at^2#
#s# is distance traveled in time #t#, #u# is initial velocity and #a# is acceleration.

Distance traveled in first #10s#
#x_1=0xx10+1/2axx10^2#
#=>x_1=50a#

Distance traveled in next #10 s#
#x_2=s_20-s_10=(0xx20+1/2axx20^2)-50a#
#=>x_2=200a-50a=150a#

Distance traveled in last #10 s#
#x_3=s_30-s_20=(0xx30+1/2axx30^2)-200a#
#=>x_3=450a-200a=250a#

the required ratio is
#x_1:x_2:x_3=50a:150a:250a#
#=>x_1:x_2:x_3=1:3:5#