On a ` 100 km ` track, a train mvoes the first ` 50 km` with a uniform speed of ` 50 km h^(-10` How fast must the train travel the next ` 50 km ` so as to have average speed ` 60 km h^(-1) for the entire trip ?

Explanation

Here, ` S_1 = 50 km , v_1 = 50 km h^(-10`
` t_1 = S-1 /v_1 = (50)/(50) = 1 h`
Let (v_2 km h^(-1)` be the uniform speed of train while voing from ` 50 km` to ` 100 km `, then time taken is ` t_2 (50) /v_2 h` Given, average speed, ` v_(av) = 60 km h^(-1)` , distance travelled `= 100 km . Therefore, total time taken is ` T= (distance travelled )/( average soeed ) = ( 100) /( 60) = 5/3 h`
Here, ` t_1 + t-2 = T`
` :. 1 + 9 50) /v_2 = 5/3 ` or ` (50)/v_2 = 5/3 - 1 = 2/`
or ` v_2 = ( 50 xx 3) /2`
`= 75 km //h`