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

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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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(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$\hat k$]