Question
The velocity of a car, in mph, is described by the equation v left parenthesis t right parenthesis equals negative t squared plus 15. Describe the acceleration of the car at 5 seconds.
(1 point)
Responses
The car is slowing down at a rate of 10 start fraction mi over h squared end fraction.
The car is slowing down at a rate of 10 Image with alt text: start fraction mi over h squared end fraction .
The car is speeding up at a rate of 10 start fraction mi over h squared end fraction.
The car is speeding up at a rate of 10 Image with alt text: start fraction mi over h squared end fraction .
The car is slowing down at a rate of 12 start fraction mi over h squared end fraction.
The car is slowing down at a rate of 12 Image with alt text: start fraction mi over h squared end fraction .
The car is speeding up at a rate of 12 start fraction mi over h squared end fraction.
(1 point)
Responses
The car is slowing down at a rate of 10 start fraction mi over h squared end fraction.
The car is slowing down at a rate of 10 Image with alt text: start fraction mi over h squared end fraction .
The car is speeding up at a rate of 10 start fraction mi over h squared end fraction.
The car is speeding up at a rate of 10 Image with alt text: start fraction mi over h squared end fraction .
The car is slowing down at a rate of 12 start fraction mi over h squared end fraction.
The car is slowing down at a rate of 12 Image with alt text: start fraction mi over h squared end fraction .
The car is speeding up at a rate of 12 start fraction mi over h squared end fraction.
Answers
GPT-4o mini
To determine the acceleration of the car at \( t = 5 \) seconds, we first need to find the expression for acceleration, which is the derivative of the velocity function.
The velocity function is given by:
\[
v(t) = -t^2 + 15
\]
We can find the acceleration \( a(t) \) by differentiating the velocity function with respect to time \( t \):
\[
a(t) = \frac{dv}{dt} = \frac{d}{dt}(-t^2 + 15) = -2t
\]
Now, we substitute \( t = 5 \) seconds to find the acceleration at that time:
\[
a(5) = -2(5) = -10 \text{ mph}^2
\]
Since acceleration is negative, it means the car is slowing down. Therefore, the car is slowing down at a rate of \( 10 \frac{\text{mi}}{\text{h}^2} \).
The correct response is:
**The car is slowing down at a rate of 10 \( \frac{\text{mi}}{\text{h}^2} \).**
The velocity function is given by:
\[
v(t) = -t^2 + 15
\]
We can find the acceleration \( a(t) \) by differentiating the velocity function with respect to time \( t \):
\[
a(t) = \frac{dv}{dt} = \frac{d}{dt}(-t^2 + 15) = -2t
\]
Now, we substitute \( t = 5 \) seconds to find the acceleration at that time:
\[
a(5) = -2(5) = -10 \text{ mph}^2
\]
Since acceleration is negative, it means the car is slowing down. Therefore, the car is slowing down at a rate of \( 10 \frac{\text{mi}}{\text{h}^2} \).
The correct response is:
**The car is slowing down at a rate of 10 \( \frac{\text{mi}}{\text{h}^2} \).**