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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 |
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Answer» (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 + 36e6t\(\hat k\) [a = -2250 sin 15t \(\hat i\) + 36 e6k\(\hat k\)] |
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