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Asked by dineshchem108 | 19 May, 2019, 09:44: PM
Position x ( in m )  =  ( t3/3 ) -3t2 +8t +4  ............................(1)

Wherer t is time in seconds

We get speed of particle dx/dt ( in m/s) by differentiating eqn.(1) as, dx/dt = t2 -6t +8   ............................(2)

We get acceleration of particle d2x/dt2 (in m/s2 ) by differentiating eqn.(2) as,  d2x/dt2 = 2t - 6  .....................(3)

for t=0 to t=5, all the above quantities are tabulated below

 t (s) x (m) displacement dx speed dx/dt (m/s) acceleration (m/s2) 0 4 0 8 -6 1 9.3 5.3 3 -4 2 10.7 1.4 0 -2 3 10 -0.7 -1 0 4 9.3 -0.7 0 2 5 10.7 1.3 3 4

Total distance travelled S1 is the sum of magnitude of all displacements from t=0 to t=5.
It is given by S1 = 0+5.3+1.4+0.7+0.7+1.3 = 9.4 m

We see from table, from t=0 to t=3, particle is moving with retardation.
Hence distance S2 travelled during retardated motion is given by, S2 = 0+5.3+1.4+0.7 = 7.4 m

Hence S1/S2 = 9.4 / 7.4 ≈ 1.3
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