A train accelerating uniformly from rest attains a maximum speed of 40 ms-1 in 20 seconds…

Q: A train accelerating uniformly form rest attains a maximum speed of 40 ms-1 in 20 seconds. It travels at this speed for 20 seconds and is brought to rest with uniform retardation in further 40 seconds. What is the average velocity during this period.

(a) 80/3 ms-1

(b) 25 ms-1

(c) 40 ms-1

(d) 30 ms-1

Ans: (b)

Sol: (i) v = u + at1

40 = 0 + a × 20 ,

a = 2 m/s2

From v2= u2 + 2 a S

402 = 0 + 2 × 2 S1

(ii) S2 =v × t1

S2 = 40 × 20

= 800 m

(iii) v = u + at

0 = 40 + a × 40 ,

a = -1 m/s2

From, v2= u2 + 2 a S

0 = 402 + 2(-1)S3

S3 = 800 m

Total distance travelled , S = S1 + S2 + S3

S = 400 + 800 + 800 = 2000 m

Total time taken = 20 + 20 + 40 = 80 s

Average velocity = 2000/80

= 25 m/s