Overtaking trains.

Let the speed of the train be x . Then, the speed of the freight train would x-20 , since it was going 20 mph slower than the train. Now, it is said that the freight train left 2 hours earlier than the train. After 2 hours, the train then started moving. Hence, the train travelled 3 hours in total, and the freight train travelled for 2+3= 5 hours . Therefore, the equation is formed:

3(x) = 5(x-20) , which is solved further as

3x = 5x-100

-2x = -100

x = -100/-2

x = 100/2

x = 50

Therefore, x (the speed of the train)= 50mph. The answer is hence 50 mph

relative velocity = 20mph. hence initial distance between them = 20*3 = 60miles. hence velocity of F is 60/2 = 30. hence velocity of P is 30+20 = 50.

lets tale speed of passenger train to be v,then speed of freight train is (v-20),the eqn would become- 3v=2(v-20)+3(v-20)

3x=5(x-20) 3x=5x-100 3x-5x= -100 -2x = -100 X= 50

As distance covered by both trains at the point of meeting is equal, If 'd' is distance covered, d1=d2 and gets cancelled we get, 5*(s-20) = 3s 5s-100 = 3s 2s-10 = 0 2s = 100 s=50 mph

Firstly consider "x" as the speed of the passenger train. That means the speed of the freight train, which is 20 mph slower, must be "x-20". The passenger train overtakes the freight train after 3 hours and also the passenger train starts 2 hours after the freight train. This means that the freight train has already traveled for 3+2= 5 hours. Because the passenger train over take after 3 hours that means the distance covered by the passenger train in 3 hours is equal to the distance covered by the freight train in 5 hours. Manipulating this info we can say : 3x= 5 (x-20) 3x= 5x-100 100 = 5x-3x 100= 2x 100/2 = x therefore, x = 50 because "x" was the speed of the passenger train, therefore the passenger train was traveling at 50 mph

let they meet X km apart from the staring point,then faster train speed=X/3,& slower train speed=X/5, X/3-X/5=20,x=150 and faster train speed=150/3=50

let the speed of passenger train be v,then speed of freight train =v-20 now,distance travelled by freight train in 5 hrs=(v-20) 5.............(1) and,same distance travelled by passenger train in 3 hrs=3v...........(2) from (1) and (2) we get, (v-20) 5=3v v=50mph

Time and speed are inversely proportional due to constant distant. Time decreases by 3/5 hence speed increases by 2/5. As we know the speed is 20mph more than the freight train's speed, x, and is 2/5 more (5/3(x)). Thus x+20=(5/3)x. Solved for x gives x=50. Answer is thus 50mph

Let the distance the both trains get until the passenger train takes over be S. Let the speed of the passenger train be v1. Let the speed of the freight train be v2. v1 = S/3 v2 = S/(3+2) =S/5 v1 - v2 = S/3 - S/5 = 2S/15 = 20 Therefore, S = 150. Therefore, v1 = 150/3 = 50