a train moves with a speed of 30 kilometre per hour in the first 15 minutes with another speed of 40 kilometre per hour in the next 15 minutes and then with a speed of 60 kilometre per hour in the last 30 minutes calculate the average
speed of the train for this journey
Answers
Answer:
47.5 kmph
Step-by-step explanation:
Distance 1 = 30×15
= 450m
Distance 2 = 40×15
= 600m
Distance 3 = 60×30
= 1800m
Total distance = 1800+600+450
= 2850m
Total time = 15 + 15+30
= 60 .
Thus ,
\begin{lgathered}average \: speed \: = \frac{2850}{60} = \frac{285}{6} \\ \\ = 47.5kmh {}^{ - 1}\end{lgathered}
averagespeed=
60
2850
=
6
285
=47.5kmh
−1
-----
Answer:47.5 km/h
Step-by-step explanation:
s1 = First constant speed
t1 = Time period of First constant speed
d1 = Distance covered by First constant speed
s2 = Second constant speed
t2 = Time period of Second constant speed
d2 = Distance covered by Second constant speed
s3 = Third constant speed
t3 = Time period of Third constant speed
d3= Distance covered by Third constant speed
s1 = 30 km/h
t1 = 15 mins (1/4 hour)
s1 = d1/t1
s1 * t1 = d1
d1 = 30 * 1/4 km
d1 = 30/4 km
d1 = 7.5 km
s2= 40 km/h
t2 = 15 mins (1/4 hour)
s2 = d2/t2
s2 * t2 = d2
d2 = 40 * 1/4 km
d2= 40/4 km
d2 = 10 km
s3= 60 km/h
t3= 30 mins (1/2 hour)
s3 = d3/t3
s3 * t3 = d3
d3 = 60 * 1/2 km
d3 = 60/2 km
d3 = 30km
Total distance = d1 + d2 + d3
= 7.5 km + 10 km + 30 km
= 47.5 km
Total time = t1 + t2 + t3
= 15 mins + 15 mins + 30 mins
= 60 mins (1 hour)
Average speed = Total distance / Total time
= 47.5 km/ 1 hour
= 47.5 km/h