Question

train moves with a constant speed of 36km
Kmb in the first 10 minutes, with another cons-
tant speed of 45 kinh in the next 10 minutes
in the
and then with an acceleration of Sms
last 10 minutes. Calculate the average speed of
the train for this journey and the total distance
travelled. (Ans. 1842 km/h: 921 km]
lony Y-axis with a​

Answer 1

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

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