If r(t) = (cos t, sin t), the acceleration is a(t) = (-cos t, -sin t).

If r(t) = (cos t, sin t), the acceleration is a(t) = (-cos t, -sin t).

• A. (-cos t, -sin t)
• B. (-sin t, cos t)
• C. (cos t, sin t)
• D. (sin t, -cos t)

Answer: (A)

Acceleration comes from differentiating the position twice with respect to t. For r(t) = (cos t, sin t), the velocity is r'(t) = (-sin t, cos t). Differentiating again gives the acceleration r''(t) = (-cos t, -sin t). So the acceleration is (-cos t, -sin t). This also fits the picture on the unit circle: the particle moves at unit speed, so its acceleration points toward the origin with magnitude 1, indeed a(t) = -r(t). The other vectors correspond to the position (cos t, sin t), the velocity (-sin t, cos t), or the negative of velocity (sin t, -cos t), which is why they don’t match the acceleration.

Acceleration comes from differentiating the position twice with respect to t. For r(t) = (cos t, sin t), the velocity is r'(t) = (-sin t, cos t). Differentiating again gives the acceleration r''(t) = (-cos t, -sin t). So the acceleration is (-cos t, -sin t).

This also fits the picture on the unit circle: the particle moves at unit speed, so its acceleration points toward the origin with magnitude 1, indeed a(t) = -r(t). The other vectors correspond to the position (cos t, sin t), the velocity (-sin t, cos t), or the negative of velocity (sin t, -cos t), which is why they don’t match the acceleration.