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 + 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 − .

Consider a particle travelling along the path given by c(t) = (−t, t, t^2). Here t denotes time. (a) Find the velocity and the speed of the particle.

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

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.

When S is measured in meters and t in seconds, find the velocity at time t =2 of the motion S = t2 – 3t2A.-4B.27C.3D.-3