The position of a particle is r(t)= (4.0t²i+ 2.4j- 5.6tk) m. (Express your answers in vector form.)
(a) Determine its velocity (in m/s) and acceleration (in m/s²) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.)
v(t)= ________m/s
a(t)= ________m/s²
(b) What are its velocity (in m/s) and acceleration (in m/s²) at time t 0?
v(0) =_______ m/s
a(0)=_______ m/s²
(a)
v(t)= [tex]8ti - 5.6k[/tex] m/s
a(t)= 8i m/s²
(b)
v(0) = -5.6k m/s
a(0)= 8i m/s²
From the question, the position of the particle is given by;
r(t)= (4.0t²i+ 2.4j- 5.6tk) -----------------(i)
(a)
(i)To get the velocity, v(t), of the particle, we'll take the first derivative of the position of the particle (given by equation (i)) with respect to time, t, as follows;
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]\frac{d(4.0t^2i + 2.4j - 5.6tk)}{dt}[/tex]
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]8ti +0j - 5.6k[/tex]
v(t) = [tex]8ti - 5.6k[/tex] --------------------(ii)
(ii) To get the acceleration, a(t), of the particle, we'll take the first derivative of the velocity of the particle (given by equation (ii)) with respect to time, t, as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex] = [tex]\frac{d(8ti - 5.6k)}{dt}[/tex]
a(t) = 8i --------------------(iii)
(b)
(i) To get the velocity of the particle at time t = 0, substitute the value of t = 0 into equation (ii) as follows;
v(t) = [tex]8ti - 5.6k[/tex]
v(0) = 8(0)i - 5.6k
v(0) = 0 - 5.6k
v(0) = -5.6k
(ii) To get the acceleration of the particle at time t = 0, substitute the value of t = 0 into equation (iii) as follows;
a(t) = 8i
a(0) = 8i
