Maths mst

Question – 3: Consider a particle moving according to the velocity function,
v(t) = 2a-3exp(-2t)+2/t+2,for t>0.

(a) If the net distance,d,covered by the particle in the time interval,[0,3],is 20,find the value of a. What is the terminal velocity of the particle?.

(b) Find the expression for the trajectory,y(t), such that, y(2) = 5 . Find the initial position of the particle.

(c) Find the acceleration, a(t), of the particle, for t > 0.

Find the acceleration when the velocity is 6.55 m/s.

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  1. The net distance (displacement) is INTEGRAL v(t) dt. Do that integration, and solve for a.

    Accelerlation is d/dt of v(t)

    y(t)= INT v (dt) . Intial position willbe the constant of integration.

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