Question

2. Each of four particles moves along an x axis. Their coordinates (in meters) as functions of time (in seconds) are given by particle 1: x(t) 3.5-2.7t particle 3: x(t) 35+2.7t? particle 2: x() 3.5+2.7 particle 4: x() 3.5-3.4t -2.7P For which of these particles is the velocity increasing for t>0?

Answer 1

To find velocity we need to differentiate x(t) with respect to t.

After that check the sign of v(t) = x'(t) on the interval 0 < t < .

If the sign is positive then velocity is increasing and vice versa.

olution: 2. Particle 1 : n(t) = 3.5-2구t3 t>0 , Monetone interval O tao Check, the bean n(t): the intewral,t 7 o Checek the sign of rce) :. Particle 2 is nerea thnovghout the intewat, t 0. Particle 3 : gelt') = 5.5 +2.4亡 t>o , > re'lt), 5.4 t → Velocity Menetone interval : o<t<

interwral , t70. >x'(t) = _ 5.4 t-5.4 1 7 the ign f ) :。Particle A is de creasing thnonahout the internal , 七70. Henee, for panttole 9-and 3 , the

3. For A-B, Ar-B and Ay B Hence, option C is correct.