Ans.

(i)

Sol. At x = 26, let t = t 0

26 = 27 t 0 –

– 27t 0 + 26 = 0

(t 0 – 1) ( + t 0 – 26) = 0

t 0 = 1, t 0 =

The earlier time gives the value of time when particle is at x = 26 m on forward journey and former time gives the value of time when particle is at x = 26 m on return journey. But in second case particle would have covered more than 26 m. So particle covers distance 26 m at

t = 1s. So velocity at this time = [27 – 3(1) 2 ] m/s

= 24 m/s.

(ii)

Sol. = – 6t

For maxima or minima, = 0

⇒ t = 0

= – 6

⇒ = – 6

So t = 0 corresponds to maxima.

v max = 27 – 3 (0) 2 = 27 m/s.

(iii) 54 m

Sol. = 27 – 3t 2 For maxima or minima,

= 0

⇒ 27 – 3t

2 = 0

t = 3 sec

∴ = – 6t

= – 18

So, t = 3 corresponds to maxima

 x max = 27 (3) – (3) 3 = 81 – 27 = 54 m