11Velocity at 4 second is
V = 9 - 4*2
V = 1m/s
Distance for 4sec - 4.5 sec
S = 0.5 - 2*0.5*0.5/2
S = 0.25m
Distance for 4.5 sec - 5sec
S = 2*0.5*0.5/2
S = 0.25
Total distance is 0.5
Option b)
A particle has a initial velocity of 9m/s due east , a constant acceleration of 2 m/s2 due west .the distance covered by it in the 5th second of its motiion is
a)0
b)0.5
c)2
d)4
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UP 0 DOWN 0 1 4
4 Answers
62This is a very good question...
the answer will seemingly be zero...
but the right way to do it is to find the velocity in the beginning of 5th second = 1 m/s east
at time =4.5 second v=0
at time = 5 second, v=1 m/s west ward
so the total distance travelled will be 0.5
1byah... i cldn get u ..........
pls make it clear fr me.....
srry if its already clear description....
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