A train moves with a constant speed of ` 36 km h^(-1)` in the first `10 minutes`, with another constant speed of ` 45km h^(-1)` in the bext ` 10 minnutes` and then with an acceleration of 5 ms^(-2)` in the last `10 minutes. Calculate the average speed of last ` 10 minutes`. Calculate the average speed of the train for this journey and the total distance travelled.

Explanation

For first ` 10 minute`, speed = 36 km h^(-1) = 10 m//s`
Didtance travelled .,
` S_1 = 10 xx ( 1=0 xx 60) = 6000 m`
distance travelled, ` S-2 = ( 25)/2 xx ( 10 xx 60) ltBrgt `= 7500 m`
For last ` 10 minutes`, ` u= ( 25//2) m//s`,
` a = 5 ms^(-1) , t = 10 xx 60 s`
Distanc etravelled, ` S_3 = ut + 1/2 at^2`
`= ( 25)/2 xx ( 10 xx 60 ) + 1/2 ( xx 60 )^2 = 90 7 500 m`
Total distance treavelled by train
` SS_1 = S_2 + S_3 = 600 = 7500 + 90 7500`
= 92 1000 m= 921 km`
Total time of joureny `= 10 + 10 + 10
` =30 min = 1/2 h`
Average speed,
` v_(av) = ( total distance travelled )/( total timne taken) = ( 921)/( 1 //2)`
` = 1 842 kmh` .