To determine the acceleration of the penguin at time t=13 seconds, we first need to find the second derivative of the displacement equation s=4cos(t).
The first derivative of s with respect to t is given by ds/dt = -4sin(t),
and the second derivative is given by d^2s/dt^2 = -4cos(t).
To find the acceleration at t=13 seconds, we substitute t=13 into the second derivative:
a = -4cos(13).
Rounding to the nearest tenth, the acceleration of the penguin at time 13 seconds is approximately -3.6 cm/sec^2. Therefore, the correct response is:
–3.6 cm/sec^2.
At time t = 0 seconds, a small toy penguin attached to the top of a spring is stretched 4 cm above its resting position. The penguin's displacement s above or below its resting position is found by the equation s=4cost . What is the acceleration of the penguin at time 13 seconds? Round your answer to the nearest tenth. (1 point) Responses –14.4 cmsec.2 – 14 . 4 cm sec. 2 –3.6 cmsec.2 – 3 . 6 cm sec. 2 –1.0 cmsec.2 – 1 . 0 cm sec. 2 3.6 cmsec.2 3 . 6 cm sec. 2 14.4 cmsec.2 14 . 4 cm sec. 2
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