Question

a particle moves in the xy plane with only an x-component of acceleration of 2 ms^-2

Answer 1

For x – bearing first, Applying the eqns. of movement, v1 = u1 + at v1 = 8i + 2i*t => (8 + 2t)I For y – heading Acceleration is zero.

In this manner speed stays consistent. v2 = u2 v2 = - 15j Thus, resultant speed of molecule at time t from conditions. v = v1 + v2 v = (8 + 2t)i – 15j m/s.

Answer 2

v = (8+2t)i - 15j m per second