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.
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.
1 answer
3 answers