Question

The velocity v, in meters per second, is given as a function of time t, in seconds, by v(t)=−0.853t^2+2.99t−8.33. What is the acceleration at time t=3.03 s?

Answer 1

To find the acceleration at time t = 3.03 s, we substitute t = 3.03 into the acceleration function: a(3.03) = -1.706(3.03) + 2.99.

Step-by-step explanation:

The acceleration of an object can be found by taking the derivative of its velocity function with respect to time. In this case, the velocity function is given as v(t) = -0.853t^2 + 2.99t - 8.33. Taking the derivative of this function will give us the acceleration function. So, a(t) = -1.706t + 2.99.

To find the acceleration at time t = 3.03 s, we substitute t = 3.03 into the acceleration function: a(3.03) = -1.706(3.03) + 2.99. Evaluating this expression will give us the acceleration at that particular time.

Answer 2

Step-by-step explanation:

To find the acceleration at a given time \( t = 3.03 \) seconds, we need to take the derivative of the velocity function with respect to time (\( t \)). The acceleration (\( a \)) is the derivative of velocity (\( v \)) with respect to time.

Given the velocity function:

\[ v(t) = -0.853t^2 + 2.99t - 8.33 \]

Let's find the acceleration (\( a \)):

\[ a(t) = \frac{dv}{dt} \]

Taking the derivative with respect to \( t \):

\[ a(t) = -1.706t + 2.99 \]

Now, substitute \( t = 3.03 \) seconds to find the acceleration at that specific time:

\[ a(3.03) = -1.706(3.03) + 2.99 \]

\[ a(3.03) \approx -1.706(3.03) + 2.99 \]

\[ a(3.03) \approx -5.169 + 2.99 \]

\[ a(3.03) \approx -2.179 \]

Therefore, the acceleration at \( t = 3.03 \) seconds is approximately \( -2.179 \) meters per second squared.