Two particles move along an x axis. The position of particla 1is given by x=6.00

Question

Two particles move along an x axis. The position of particla 1is given by x=6.00t2+3.00t+2.00 (in meters and seconds);the acceleration of particle 2 is given by a=-8.00t (in meters perseconds squared and seconds) and, at t=0, its velocity is 20m/s.When the velocities of the particles match, what is theirvelocity? Two particles move along an x axis. The position of particla 1is given by x=6.00t2+3.00t+2.00 (in meters and seconds);the acceleration of particle 2 is given by a=-8.00t (in meters perseconds squared and seconds) and, at t=0, its velocity is 20m/s.When the velocities of the particles match, what is theirvelocity?

Explanation / Answer

Ok, the position of particle 1 is given by x = 6.t² + 3.t + 2 so that its velocity at time t is v1(t) = dx/dt = 12.t + 3. Particle 2 has an initial velocity v2(0) of 20 m/s, and anacceleration a of - 8 m/s², so that its velocity at time tis v2(t) = v2(0) - a.t = 20 - 8.t The two velocities will be equal when v1(t) = v2(t) or 12.t + 3 = 20 - 8.t which gives 20.t = 18 that is, at a time t = 0.9 s. Their common velocity is then (12 * 0.9 + 3) = (20 - 8 * 0.9) =13.8 m/s.

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