Question

this question is designed to be answered without a calculator.
a particle travels along the x-axis so that its position, in centimeters, after t seconds is given by the function ( x(t) = -4t^3 - 8t^2 + 16t - 64 ). what is the acceleration of the particle at time ( t = 1 ) sec?
( \bigcirc ) ( -52 ) cm/sec(^2)
( \bigcirc ) ( -40 ) cm/sec(^2)
( \bigcirc ) ( -24 ) cm/sec(^2)
( \bigcirc ) ( -12 ) cm/sec(^2)

Explanation:

Step1: Find the velocity function (first derivative)

The position function is \( x(t) = -4t^3 - 8t^2 + 16t - 64 \). The velocity \( v(t) \) is the first derivative of \( x(t) \). Using the power rule \( \frac{d}{dt}(t^n)=nt^{n - 1} \), we get:
\( v(t)=x^\prime(t)=\frac{d}{dt}(-4t^3 - 8t^2 + 16t - 64)=-12t^2-16t + 16 \)

Step2: Find the acceleration function (second derivative)

The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \). Differentiating \( v(t)=-12t^2-16t + 16 \) with respect to \( t \):
\( a(t)=v^\prime(t)=\frac{d}{dt}(-12t^2-16t + 16)=-24t-16 \)

Step3: Evaluate acceleration at \( t = 1 \)

Substitute \( t = 1 \) into the acceleration function \( a(t) \):
\( a(1)=-24(1)-16=-24 - 16=-40 \)

Answer:

\(-40\space cm/sec^2\) (corresponding to the option: \(-40\space cm/sec^2\))