Acceleration 2

**Sunyata** · May 5th 2010, 07:18 AM

The velocity of a particle starting at the origin is v = (cos2x)^2. Show that acceleration = -4(cos2x)^3 sin2x. When I differentiate v to get acceleration, i do not get the cubed part. Please help???

**skeeter** · May 5th 2010, 09:22 AM

Originally Posted by Sunyata

The velocity of a particle starting at the origin is v = (cos2x)^2. Show that acceleration = -4(cos2x)^3 sin2x. When I differentiate v to get acceleration, i do not get the cubed part. Please help???

$a = \frac{d}{dt} [\cos(2x)]^2$

$a = 2[\cos(2x)] \cdot [-\sin(2x)] \cdot 2 \cdot \frac{dx}{dt}$

$<br /> a = 2[\cos(2x)] \cdot [-\sin(2x)] \cdot 2 \cdot \cos^2(2x)<br />$

simplify ...

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