In one dimensional motion position of a particle from a fixed point is (say P ) is +2m at time t=0 and the particle is at P at t=10 s. If velocity of particle is zero at t=6 s, determine the constant acceleration a and velocity at t=10 s.
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a
1.8m/s
b
0.9m/s
c
0.8m/s
d
0.1m/s
answer is C.
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Detailed Solution
s=s0+ut+12at2 at t=0,s=s0=2 m ----(i) at t=10 s s=0=s0+ut+12at2 0=2+10u+50a 10u+50a=-2 v=u+a t ----(ii) 0=u+6a u+6a=0 ----(iii) Solving Eqs. (ii) and (iii), we get u=-1.2 m/s and a=0.2 m/s2 Now, v=u+at v=(-1.2)+(0.2)(10) =0.8m/s