Question

When two body approach each other with different speeds, the distance between them decreases by 120m for every one minute.if they are moving in direction,the distance between them increases by 90m for every one minute. The speeds of the bodies are

Answer 1

Answer:

Suppose both start at the origin

x1 = V1 t       distance 1 travels

x2 = V2 t     distance 2 travels

(x2 - x1) / t = V2 - V1 = 90/60 = 1.5 m/s    both in same direction

(x2 + x1) / t = V2 + V1 = 120 / 60 = 2 m/s    moving in opposite directions

adding equations  2 V2 = 3.5 m/s

so V2 = 1.75 m/s   and V1 = .25 m/s

Note: in the second part the relative speed of approach will equal

the relative speed of separation - you could write

x2 = A - V2 t and have x1 start at the origin but the A will cancel when you subtract the equations

Suppose 1 starts at x1 = 0 and 2 starts at x2 = 200  and they are moving in opposite directions

After 60 sec  x1 is at 15 m   and  x2 is at 200 - 105 = 95 m

Their separation has decreased by 120 m as specified.

Answer 2

Answer:

Suppose both start at the origin

x1 = V1 t       distance 1 travels

x2 = V2 t     distance 2 travels

(x2 - x1) / t = V2 - V1 = 90/60 = 1.5 m/s    both in same direction

(x2 + x1) / t = V2 + V1 = 120 / 60 = 2 m/s    moving in opposite directions

adding equations  2 V2 = 3.5 m/s

so V2 = 1.75 m/s   and V1 = .25 m/s

Note: in the second part the relative speed of approach will equal

the relative speed of separation - you could write

x2 = A - V2 t and have x1 start at the origin but the A will cancel when you subtract the equations

Suppose 1 starts at x1 = 0 and 2 starts at x2 = 200  and they are moving in opposite directions

After 60 sec  x1 is at 15 m   and  x2 is at 200 - 105 = 95 m

Their separation has decreased by 120 m as specified.