What if the acceleration is not constant? A particle starts from the origin with

Question

What if the acceleration is not constant? A particle starts from the origin with velocity 2 m/s at t = 0 and moves in the xy plane with a varying acceleration given by = (9?t) m/s2, where t is in s. Answer the following using t as t as necessary.
(a) Determine the vector velocity of the particle as a function of time.
(t) = (

+

) m/s

(b) Determine the position of the particle as a function of time.
(t) = (

+

) m

Explanation / Answer

For answering this question You have to give the acceleration in the form similar to those of v(t) and r(t), i.e. a(t) = (??? i + ??? j ) = (ax*i + ay*j) m/s2 vx = dx/dt ==> x = ?vx dt vy = dy/dt ==> y = ?vy dt ax = d(vx)/dt ==> vx = ?ax dt ay = d(vy)/dt ==> vy = ?ay dt Let a = A(8vt)i +B(8vt)j = ax*i+ay*j, where A=cosa and B=sina, a is the angle between vector a and +x axis. Then vx = ?ax dt = ?A(8vt) dt = (2A/3)*8t^(3/2) + C = (16A/3)*t^(3/2) + C vx(0) = 4 ==> C = 4 ==> vx = 4 + (16A/3)*t^(3/2) vy = ?ay dt = ?B(8vt) dt = (2B/3)*8t^(3/2) + C = (16B/3)*t^(3/2) + C vy(0) = 0 ==> C = 0 ==> vy=(16B/3)*t^(3/2) v(t) = [4 + (16A/3)*t^(3/2)]i + [(16B/3)*t^(3/2)]j r = x*i + y*j x = ?vx dt = ? [4 + (16A/3)*t^(3/2)] dt = 4t + (2/5)*(16A/3)*t^(5/2) + C = 4t + (32A/15)*t^(5/2) +C x(0) = 0 ==> C = 0 ==> x = 4t + (32A/15)*t^(5/2) Similarly y = (32B/15)*t^(5/2) r(t) = [4t + (32A/15)*t^(5/2)]i + [(32B/15)*t^(5/2)]j

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