Asked by 1145003
The velocity of a car, in mph, is described by the equation v(t) = -t^2 + 15. Describe the acceleration of the car at 5 seconds.
A. The car is slowing down at a rate of 10 mi/h^2
B. The car is speeding up at a rate of 10 mi/h^2
C. The car is slowing down at a rate of 12 mi/h^2
D. The car is speeding up at a rate of 12 mi/h^2
A. The car is slowing down at a rate of 10 mi/h^2
B. The car is speeding up at a rate of 10 mi/h^2
C. The car is slowing down at a rate of 12 mi/h^2
D. The car is speeding up at a rate of 12 mi/h^2
Answers
Answered by
GPT-4o mini
To determine the acceleration of the car at a specific time using the velocity equation \( v(t) = -t^2 + 15 \), we first need to find the acceleration, which is the derivative of the velocity function with respect to time.
1. **Differentiate the velocity function**:
\[
a(t) = \frac{dv}{dt} = \frac{d}{dt} (-t^2 + 15) = -2t
\]
2. **Calculate the acceleration at \( t = 5 \) seconds**:
\[
a(5) = -2(5) = -10 \text{ mi/h}^2
\]
The negative sign indicates that the acceleration is in the opposite direction of the velocity, suggesting that the car is slowing down.
Thus, at 5 seconds, the car is slowing down at a rate of 10 mi/h².
The correct answer is:
**A. The car is slowing down at a rate of 10 mi/h².**
1. **Differentiate the velocity function**:
\[
a(t) = \frac{dv}{dt} = \frac{d}{dt} (-t^2 + 15) = -2t
\]
2. **Calculate the acceleration at \( t = 5 \) seconds**:
\[
a(5) = -2(5) = -10 \text{ mi/h}^2
\]
The negative sign indicates that the acceleration is in the opposite direction of the velocity, suggesting that the car is slowing down.
Thus, at 5 seconds, the car is slowing down at a rate of 10 mi/h².
The correct answer is:
**A. The car is slowing down at a rate of 10 mi/h².**