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Motion In A Plane

Question
CBSEENPH11020756

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms-1 to 20 ms-1 while passing through a distance 135 m in t second.  The value of t is 

  • 10

  • 1.8

  • 12

  • 9

Solution

D.

9

The problem requires kinematics equations of motion
Let u and v be the first and final velocities of particle an a and s be the constant acceleration and distance covered by it. 
From third equation of motion
v2 = u2 + 2as
(20)2 = (10)2 + 2a x 135
or
a = 300/2 x135 = 10/9 ms-2
Now using first equation of motion,


v= u +at
or 
straight t space equals space fraction numerator straight v minus straight u over denominator straight a end fraction space equals space fraction numerator 20 minus 10 over denominator left parenthesis 10 divided by 9 right parenthesis end fraction space equals space fraction numerator 10 space straight x 9 over denominator 10 end fraction space equals space 9 straight s