See question which is attached. Particle P moves along the y-axis so that its position at time t is given byy(t) = 4t²5t + 7 for all times t. A second particle, Q, moves along the x axis so that itssin(at)position at time t is given by x(t)for all times t # 2.2-t=a. Find lim x(t). Show the work that leads to your answer.t→2b. Show that the velocity of particle Q is given by vo(t)=2π сos(лt)-лt cos(πt)+sin(πt)(2-t)²for alltimes t # 2.c. Use calculus and algebra to find a value of t such that P and Q reach the same velocity.Show the work that leads to your answer. State the velocity.

Particle P move along y-axis, and its position vector is given by, y(t)=4t2-5t+7. Particle Q move along x-axis and its position vector is given by, x(t)=sinπt2-t. (a) To Find:limx→2x(t). (b) To Find: Velocity of the particle Q . (c) To Find: The time when velocity of both particles are same.,Particle P move along y-axis, and its position vector is given by, y(t)=4t2-5t+7. Particle Q move along x-axis and its position vector is given by, x(t)=sinπt2-t. (a) limt→2x(t)=sinπt2-t.....................00 form=πcosπt-1..............Lo^hpital rule=-π thus, limx→2x(t)=-π.(b) Velocity along x-axis of particle Q, vQ(t)=dxdt where x(t)=sinπt2-t. dxdt=2-tπcosπt+sinπt2-t2=2πcosπt-πtcosπt+sinπt2-t2 Therefore, vQ(t)=2πcosπt-πtcosπt+sinπt2-t2. (c) Velocity of particle P, vP(t)=dydt where y(t)=4t2-5t+7. vP(t)=8t-5. velocity of both particles are same when, vP(t)=vQ(t). 2πcosπt-πtcosπt+sinπt2-t2=8t-5. Using calculator, t=0.599. Thus, At t=0.599 velocities of both particles are same.