Question

Suppose the position of an object is given by r = (3.0t2 i – 6.0t3 j) m. [DIFFERENTIATION] 

1. Determine its velocity, v, and acceleration, a, as a function of time. [ANS v = (6.0t i – 18.0t2 j) m/s, a = (6.0 i – 36.0t j) m/s2] 
2. Determine r and v at time t = 2.5 s. [ANS r = (19i – 94j) m, v = (15i – 110j) m/s]  

Answer 1

1)

$$v=\dfrac{dr}{dt}=\dfrac{d}{dt}[(3.0t^2\ i-6.0t^3\ j)\ m]=(6.0t\ i-18.0t^2\ j)\ \dfrac{m}{s},$$ $$a=\dfrac{dv}{dt}=\dfrac{d}{dt}[(6.0t\ i-18.0t^2\ j)\ m]=(6.0\ i-36.0t\ j)\ \dfrac{m}{s^2}.$$

2)

$$r(t=2.5\ s)=(3.0\times(2.5\ s)^2\ i-6.0\times(2.5\ s)^3\ j)\ m,$$ $$r(t=2.5\ s)=(19.0i-94j)\ m.$$ $$v(t=2.5\ s)=(6.0\times2.5\ s\ i-18.0\times(2.5\ s)^2\ j)\ \dfrac{m}{s},$$ $$v(t=2.5\ s)=(15i-112j)\ \dfrac{m}{s}.$$