Question

Find the velocity and acceleration of the point described by the following position vectors (t = time in seconds); 

(a) r = 16t i + 25t² j+ 33 k cm. 

(b) r = 10 sin15t i + 35tj + e^(6t) k cm

Answer 1

(a) Given

 \(\vec r=16t\hat i+25t^2\hat j+33\hat k\) cm

\(\vec r=0.16t\hat i+.25t^2\hat j+.33\hat k\) m

Velocity v = \(\frac{d\vec r}{dt}\)

= \(\frac{d}{dt}(0.16t\hat i+.25t^2\hat j+.33\hat k)\) 

\(\vec v=0.16\hat i+0.50t\hat j+0\) 

\(\vec v=0.16\hat i+0.50t\hat j\)

\(\hat v=0.16\hat i+0.50t\hat j\) m/s

acceleration

a = \(\frac{dv}{dt}\)

a = \(\frac{d}{dt}(0.16\hat i+0.50t\hat j)\) 

a = 0 + 0.50 j

a = 0.50j m/s²

(b) \(\vec r=10sin15t\hat i+35t\hat j+e^{6t}\hat k\) 

Velocity v = \(\frac{d\vec r}{dt}\)⇒ \(\frac{d}{dt}(10sin15t\hat i+35t\hat j+e^{6t}\hat k)\) 

\(\vec V=10cos15t\times15\hat i+35\hat j+e^{6t}.6\hat k\)

\([\vec v=150cos15t\hat i+35\hat j+6e^{6t}\hat k]\) cm/sec

acceleration a = \(\frac{d\vec v}{dt}\)

a = \(\frac{d}{dt}(15 cos15t\hat i+35\hat j+6e^{6t}\hat k)\) 

a = 150(-sin15t)15\(\hat i\) + 0 + 36e^6t\(\hat k\)

[a = -2250 sin 15t \(\hat i\) + 36 e^6k\(\hat k\)]