Question

7. The acceleration of a particle moving along the x-axis at time t is given by a(t) = 6t – 11. If the velocity is 10 when t = 0 and the position is 4 when t=0, then the particle is changing direction at what time(s)?

Answer 1

Given, acceleration, a (t) = 6t-ll. Velocity, Ult) = f act).dt V(E) = 6.4 11t+C V(t) = 3 1² 11t+G Given, at t=0, V=10 = V(o)=10. 10=0-0th, (4=10] For So, V(t) = 3:(+²) 1lt to. Particle to cheupe its direction, Velocity, V(t) =0 and is placement (4) > 3 t²_11t to=0 act) <0 * 342_66-5t to co -maximum. or minimum, face) <0 [alt so → 32(x-2)-5. (E-2) =0 t=2 or t = 5/5 So, a(z) – 6(2)=11 = 1 >0 = $(2) > Minimum. =-120 = (3/3) - Maximum. the particle is changing its direction and also at t = 2, The particle is changing its direction