The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 4,    v(0) = −32,    0 ≤ t ≤ 6Exercise (a)Find the velocity at time t.Step 1The velocity function is the antiderivative of the acceleration.v(t) = (2t + 4) dt = $$t2+4t + CStep 2We must determine the value of C. We know that v(0) = −32.Substituting 0 into our antiderivative gives −32 = v(0) = 3 + C. Therefore, C = . Exercise (b)Find the distance traveled during the given time interval.Step 1The velocity function is v(t) = t2 + 4t − 32, and so the distance traveled in the time interval 0 ≤ t ≤ 6 is given by 6|t2 + 4t − 32| dt0.Remembering that |z| =   z ≥ 0   z < 0, we must determine where v(t) = t2 + 4t − 32 is positive or negative.v(t) can be factored as t2 + 4t − 32 = t + t − .

Find the velocity at time t.Step 1The velocity function is the antiderivative of the acceleration.v(t) = (2t + 4) dt = $$t2+4t + CStep 2We must determine the value of C. We know that v(0) = −32.Substituting 0 into our antiderivative gives −32 = v(0) =

The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.a(t) = 2t + 2,    v(0) = −15,    0 ≤ t ≤ 5(a) Find the velocity at time t.v(t) = m/s(b) Find the distance traveled during the given time interval.

particle moves along a straight line with an acceleration described by equation a=-8s^-2 where a is in m/sec^2 and s in meter. When t= 1 sec, s= 4 m and v=2 m/sec. Determine the acceleration when t = 2 seconds

A particle moves along a straight line such that its displacement at any time t is given by S = t3 – 6t2 + 3t + 4 metres. The velocity when the acceleration is zero is :4 ms–1– 12 ms–1 42 ms–1– 9 ms–1

11 A particle travels in a straight line so that, t seconds after passing a fixed point A on the line, itsacceleration, a ms–2, is given by a = –2 – 2t. It comes to rest at a point B when t = 4.(i) Find the velocity of the particle at A