u = 48 m/sec a = – 10 m/s 2

so, by v = u + at 0 = 48 – 10 t

so, t = 4.8 s

this means that the particle comes to rest at t = 4.8 s and turns back covering some distance backwards for rest of the motion.

for the forward journey distance travelled in last 0.8 second before stopping and returning will be

(s 4.8 – s 4 ) where, s 4.8 and s 4 are distances travelled in 4.8 seconds and 4 seconds respectively.

s 4.8 = 48 × 4.8 + × –10 × 4.8 2 = 48 × 2.4

s 4 = 48 × 4 + × – 10 × 4 2 = 16 × 7

(s 4.8 – s 4 ) = (48 × 2.4) – (16 × 7)

Distance travelled 0.2 s during backward journey = s 0.2 = × 10 × 0.2 2 = 0.2

So, total distance travelled = (48 × 2.4) – (16 × 7) + 0.2 = m.