A bicycle moves with a constant velocity of+6.0m/s and moves with acceleration of-20m/s^2. Show that after 6 s the particle will be t starting point.
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calculate time taken for bicycle to be at rest i.e v=0, use v = u + at
t1 = 6/20 sec, use v^2-u^2 = 2*a*s to calculate s as well => s = 36/40
Now the particle is coming back to same point with so s = -36/40 with u = 0
use s = ut+1/2 at^2 => -36/40 = 1/2*-20*t^2 or t2 = sqrt (36/400) = 6/20 sec
time taken is t1+t2 = 12/20 sec (I feel that your a = -20 is wrong, it should be -2 m/sec^2 to get 6s as answer)
Other method - Direct
use s = ut-1/2at^2 where s= 0, u = 6 and a =-2 (considering the figure in que is wrong)
0 = 6*t-1/2*2*t^2 or t^2-6t = 0 or t = 6 sec or t = 0 sec
t1 = 6/20 sec, use v^2-u^2 = 2*a*s to calculate s as well => s = 36/40
Now the particle is coming back to same point with so s = -36/40 with u = 0
use s = ut+1/2 at^2 => -36/40 = 1/2*-20*t^2 or t2 = sqrt (36/400) = 6/20 sec
time taken is t1+t2 = 12/20 sec (I feel that your a = -20 is wrong, it should be -2 m/sec^2 to get 6s as answer)
Other method - Direct
use s = ut-1/2at^2 where s= 0, u = 6 and a =-2 (considering the figure in que is wrong)
0 = 6*t-1/2*2*t^2 or t^2-6t = 0 or t = 6 sec or t = 0 sec
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