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.

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.

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.b) Find the unit tangent vector T(t), the principal normal vector N(t), the binormal vector B(t), the curvature κ(t) and the torsion τ (t) at time t.

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

A particle moves along a line with a velocity v(t)=t2+3t−4, measured in meters per second. Find the total distance the particle travels from t=0 seconds to t=2 seconds.Enter an exact answer.Provide your answer below:

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 − .