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

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.

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 position of an object moving along x-axis is given by x = a + bt2 where a = 8.5 m, b =2.5 m s–2 and t is measured in seconds. What is its velocity at t = 0 s and t = 2.0 s. Whatis the average velocity between t = 2.0 s and t = 4.0 s ?

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 − .

The velocity of a particle is given by the equationv = 9t2 i + t3 j + (0.75t + 3) k m/swhere t is in seconds. Determine the following:(a) the magnitude of the particle’s acceleration when t=1.75 seconds, a = Answer m/s2.(b) the magnitude of the particle's displacement when t=1.75 seconds (r=0 when t=0), r = Answer metres.