Question

At t=2s car a is at x=300m and car b is at x=-50m. car A is...

At t=2s car a is at x=300m and car b is at x=-50m. car A is traveling at a constant velocity of 15 m/s in the positive x direction. car B is traveling at a constant velocity of 20 m/s, also in the positive x direction. where would car B collide with car A if they both stayed on course. Please explain every step.

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Answer #1

Since both are moving in same direction therfore their relative velocity will be
Vr = VB - VA
where VB is velocity of car B = 20 m/s
VA is velocity of car A = 15 m/s
Vr = 20 -15 = 5 m/s
The distance between the two car at (t = 2s) S=350 m as shown in the figure.
Now this distance has to be cover with relative speed so they can collide
time = distance / relative speed
= S/Vr = 350/5 = 70 s
Hence the time at collision = initial time + time taken to collision = 2 + 70 = 72 s
Therfore the car will collide at 72 s
Now if we have to find the distance covered by car A in 70 s
SA = speed*time = 15*70 = 1050 m
Therefore the collision will takes place = 300 + 1050 = 1350 m
Hence collision will takes place at x = 1350 m and at time t = 72 s

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