- #1
Hijaz Aslam
- 66
- 1
Q. Two particles start simultaneously from the same point and move along two straight lines, one (particle A) with uniform velocity v and other (particle B) with a uniform acceleration a. If ##\alpha## is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given by:
(a) ##vsin\alpha/a## (b) ##vcos\alpha/a## (c) ##vtan\alpha/a## (d) ##vcot\alpha//a##
My text gives the answer as (b) ##vcos\alpha/a## . I think they reached at the answer as follows:
At an instant (the instant when the relative velocity is minimum; let's say) V=at, therefore ##vcos\alpha-V=0## where 0 is the shortest relative velocity (assumed). Therefore ##vcos\alpha=at## or ##t=vcos\alpha/a##.
But this answer seems unconvincing. It is not necessary that for the relative velocity to be minimum the component of velocity of the 'A' (which attains a constant velocity) should be equal to the velocity of the 'B' (which has the constant acceleration) at a particular instant. Can anyone provide the precise solution?
(a) ##vsin\alpha/a## (b) ##vcos\alpha/a## (c) ##vtan\alpha/a## (d) ##vcot\alpha//a##
My text gives the answer as (b) ##vcos\alpha/a## . I think they reached at the answer as follows:
At an instant (the instant when the relative velocity is minimum; let's say) V=at, therefore ##vcos\alpha-V=0## where 0 is the shortest relative velocity (assumed). Therefore ##vcos\alpha=at## or ##t=vcos\alpha/a##.
But this answer seems unconvincing. It is not necessary that for the relative velocity to be minimum the component of velocity of the 'A' (which attains a constant velocity) should be equal to the velocity of the 'B' (which has the constant acceleration) at a particular instant. Can anyone provide the precise solution?