Question

A train moves with a constant speed of 36km hr. in the first 10 minute, with another constant speed 45km hr. in the next 10 minutes and then with an acceleration of 5ms in the last 10 minutes. Calculate the average speed of the train for this journey and the total distance travelled​

Answer 1

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Answer:

distance travelled is 921km

Explanation:

speed=36km/h

36×5/18=10m/s

distance=v×t

10×(10×60)=6000m

2nd speed=45km/h

45×5/18=25/2 m/s

distance=25/2×600=7500m

3rd a=5m/s^2

t=600m

v=25m/sec

s=ut+1/2×at^2

=25/2×600+1/2×5×(600)^2

=7500+900000

=907500m

Adding total distance=6000+7500+907500

=921000/1000km

=921km

Answer 2

Answer:

here's your answer buddy..

Explanation:

1.  speed= 36km/h or 10m/s

    time=10min or 600sec

    dist=speed * time=10*600=6000m

2.  speed= 45km/h or 25/2 m/s

    time=10min or 600sec

     dist= 25/2*600=7500m

3. acc. to second law of motion

   s=ut+1/2a$t^{2}$

   s= 25/2 *600+ 1/2*5*600*600

   s=7500+900000

   s=907500m

So, total dist=6000+7500+907500

                     =921000m or 921 km

     average speed= dist/time =921000/10+10+10*60

                                                  =511.6m/s

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