Question

3. An express train starts from rest and accelerates
uniformly at a rate of 1 ms 2 for 20 s. It then maintains
a constant speed for 120 s. Then, the driver applies
brakes and the train comes to rest in 10 s. Calculate
the (a) maximum velocity of the train, (b) retardation
on applying brakes, and (c) total distance travelled by
the train.
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Answer 1

Answer:17.69 m/s

Explanation:

(i) For the first 10 s, initial velocity u = 0

Acceleration a = 2 m/s2

Time taken t = 10 s

Let ‘v’ be the maximum velocity reached.

Using the first equation of motion

v = u + at

We get

V = (0) + (2) (10) = 20 m/s

(ii) For the last 50 s: Final velocity = 0 m/s, initial velocity = 20 m/s.

Acceleration = (Final velocity – Initial velocity)/time

= (0 – 20)/50 = -0.4 m/s2

Retardation = 0.4 m/s2

(iii) Total distance travelled = Distance travelled in the first 10 s + Distance travelled in 200 s + Distance travelled in last 50 s

Distance travelled in first 10s (s1) = ut + (1/2) at2

S1= (0) + (1/2) (2) (10)2

S1= 100 m

Distance travelled in 200s (s2) = speed × time

S2 = (20) (200) = 4000 m

Distance travelled in the lastinlast 50s (s3)= ut+(1/2)at^2

Here, u = 20 m/s, t = 50 s and a = -0.4 m/s2

S3= (20)(50) + (1/2) (-0.4) (50)2

S3= 1000 – 500

S3= 500 m

Therefore, total distance travelled = S1 + S2 + S3 = 100 + 4000 + 500 = 4600 m

(iv) Average velocity = Total distance travelled/total time taken

= (4600/260) m/s

= 17.69 m/s