A particle starts from rest at t=0 and undergoes an acceleration as shown in figure.
Draw the displacement-time graph of the motion of the particle during the same time interval. In both the above cases, explain the various steps.

Explanation

The displacement of the particle in time interval t = 0 to t = 2 is given by
`s=ut+(1)/(2)" at"^(2) =0+(1)/(2) xx3xxt^(2) (3)/(2) t^(2)`
Therefore, the displacement at t = 0, t = 1 s and t = 2 s respectively are 0, 1.5 m and 6 m. In the time interval from t = 2s to t = 4 or t . = 0 or t . = 2s ( t . = t – 2), the displacement is given by
`s.=ut.+(1)/(2) a(t.)^(2) =6t. (1)/(2) xx3xx t^(2) =6t-1.5 t^(2)`
(Hence for this time interval `u = 6 m//s and a = – 3 m//s^(2))`. The displacement at t . = 1s and t . = 2 i.e., t = 3s and t = 4s are, therefore, (6 –1.5) = 4.5 m and (12 – 6) = 6 m respectively more than the displacement at t = 2s which is 6 m . Hence the displacement at t = 3s and t = 4s are (4.5 + 6) = 10.5 cm and (6 + 6) = 12 m respectively.
These results are tabulated below:
Time (s) 0 1 2 3 4
Displacement (m/s) 0 1.5 6 10.5 12
Figure shows the displacement-time graph of the motion