Question

10. (a) A train accelerated from 20 km/hr to 80km/hr in 4 minutes. How much distance does it cover in this period? Assume that the tracks are straight.
(b) A train is travelling with a velocity of 45km/hr. Calculate the distance travelled by it in 1hr, 1 min, 1 sec.

Answer 1

Answer:

A)

Given :-

Time = 4 minutes = 4/60 hours = 1/15 hours.

Intital Velocity(u) = 20 km/hr

Final Velocity (v) = 80km/hr

for finding Acceleration (a) :-

Use first equation of motion,

v = u + at

⇒ a=v-u/t

⇒a = 80-20/1/15

⇒a = 60 × 15

⇒a = 900km/hr ²

For finding (s) distance :-

Use second equation of motion

s=ut+at²/2

⇒s =20(1/15)+900(1/15)(1/15)/2

⇒s =4/3+2

⇒s =10/3km

Therefore, The distance it covers in this period is 10/3 km

B)

First case :-

Speed = 45 km/hr

Time = 1 hr

Distance = speed x time

= 45 x 1

= 45 km ( 45000 m)

Therefore, the train travels 45 km in 1 hour.

Second case :-

Speed = 45 km/hr or we can also take as

45000 m/ 60 min

Time = 1 min

Distance = speed x time

=( 45000/60 ) x 1

= 45000 / 60

= 750 metres

Therefore, the train travels 750 m in 1 minute.

Third case :-

Speed = 45000 m/ 60 min (or )

45000 m/ 3600sec

Time = 1 second

Distance = speed x time

= (45000 / 3600) x 1

= 45000 / 3600

= 12.5 metres

Therefore, the train travels 12.5 m in 1 second.