Physics problem (relative velocity) ?

Question

If you are walking at constant velocity of 5 km/h and a car passed you by at the speed of 20 km/h from behind, what is the car's velocity from your viewpoint?

explanation needed !

Answer 1

according to relative velocity addition formulae ->

u'=u-v/1-uv/c^2

here u=20 v=5

u'=20-5/1 = 15 m/s since both speed are less so it will not make any measurable change in relative velocity .......

Answer 2

If that car ran into you, how fast would it be coming at you?

It's coming 20 kph at you and you're walking 5 kph at it.  Thus, you are closing the car at 20 + 5 = 25 kph aren't you?  This results because the two veolcities, yours and its, are additive.

Now turn around...it's still coming at you at 20 kph.  But now you're walking in a direction away from the car, still at 5 kph, but facing away.  Thus, the car is approaching from behind and the rate of closure is now 20 - 5 = 15 kph.

And there you have it.  Velocities are additive (and subtractive).  When the velocities face each other they add up; when they face the same direction, they subtract.

The whole point of this problem is to point out that velocities have direction as well as magnitude.  And that direction has to be taken into account when adding or subtracting them.

Answer 3

The car is moving at 15 km/h relative to you, because it's moving 15 km/h faster than you in the same direction.  If you were running at 20 km/h (ok, riding a bike), the car would be moving at the same velocity as you and would appear to be standing still from your point of view, or have a relative velocity of 0.