The two vehicles are moving one after the other in the same direction. At the initial time the distance between vehicles is 300 m. The velocity of the first vehicle is 10 m/s and the velocity of the second vehicle is twice larger than the velocity of the first vehicle. The second vehicle is behind the first vehicle. The both vehicles simultaneously accelerate with constant acceleration. The acceleration of the first vehicle equals 1 m/s^2, and the acceleration of the second vehicle in two times less. What is the minimum distance between the vehicles and at what time does the minimum distance occur?

Solution. Posted on Sunday, November 9, 2014.

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The vehicles will be at the minimum distance when their velocities are the same.

Applying v = v0 + a*t

v2 = 20msec-1 + .5msec-2 * t = 10 msec-1 + 1msec-2 * t = v1

solving for t, t = 20 secs.

To get the minimum distance between the two vechicles, apply s = s0 + v0*t + .5*a*t^2 where t = 20 secs then subtract the two distances.

s2 = 0 + 20msec-1*20sec + .5 * .5msec^(-2) * 400sec^2

s2 = 500 m

s1 = 300m + 10msec-1*20sec + .5*1msec^(-2) * 400sec^2

s1 = 700 m

Minimum distance = s1 – s2 = 200 m

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Thank you for your solution !

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