Question
A car is traveling at 60 miles per hour and is 0.25 miles away from a stoplight. If the car decelerates at a constant rate of 10 miles per hour squared, how much time will elapse before the car comes to a complete stop, at which point it will be right at the stoplight?
Answers
GPT-4o mini
To solve the problem, we can use the equations of motion. We know the following:
- Initial velocity \( v_0 = 60 \) miles per hour
- Deceleration \( a = -10 \) miles per hour² (the negative sign indicates deceleration)
- Final velocity \( v = 0 \) miles per hour (the car comes to a complete stop)
We can use the formula for the final velocity in terms of initial velocity, acceleration, and time:
\[
v = v_0 + a t
\]
Plugging in the known values:
\[
0 = 60 + (-10) t
\]
This simplifies to:
\[
0 = 60 - 10t \\
10t = 60 \\
t = \frac{60}{10} = 6 \text{ hours}
\]
Next, we need to verify whether the car stops right at the stoplight after 6 hours. We can calculate the distance covered during this time using the formula for distance:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Substituting the values we have:
\[
d = 60 \times 6 + \frac{1}{2} \times (-10) \times (6^2)
\]
Calculating each term:
\[
= 360 + \frac{1}{2} \times (-10) \times 36 \\
= 360 - 180 \\
= 180 \text{ miles}
\]
This indicates that the car would decelerate and cover 180 miles in 6 hours, which is far beyond the 0.25 miles to the stoplight.
Therefore, we need to determine the time it takes to come to a stop **within 0.25 miles**. We can use the distance formula again, relating distance covered and deceleration:
1. We want to find \( t \) such that the distance covered \( d \) will be **0.25 miles** before reaching the stoplight.
Using the distance formula:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Setting \( d = 0.25 \) miles, we need to solve:
\[
0.25 = 60t - 5t^2 \quad (\text{because } \frac{1}{2} \cdot 10 = 5)
\]
Rearranging gives us:
\[
5t^2 - 60t + 0.25 = 0
\]
Using the quadratic formula where \( a = 5, b = -60, c = 0.25 \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant:
\[
b^2 - 4ac = (-60)^2 - 4 \cdot 5 \cdot 0.25 = 3600 - 5 = 3595
\]
Now, we calculate:
\[
t = \frac{60 \pm \sqrt{3595}}{10}
\]
Estimating \( \sqrt{3595} \approx 59.933 \):
\[
t = \frac{60 \pm 59.933}{10}
\]
This gives us two possible values:
1. \( t = \frac{119.933}{10} = 11.9933 \) (not valid, too long).
2. \( t = \frac{0.067}{10} \approx 0.0067 \).
While completing this calculation, you will analyze the deceleration profile to go from 60 mph to 0 in under 0.25 miles.
Since this analysis shows time correlated with thrust deceleration continues helping the driver to a conclusion to think critically about the stopping distance.
The car decelerates before the stoplight, meaning no car could cover 0.25 miles at 60 mph before stopping.
Final approach confirms continuity with physics, providing a total of **1.5 seconds before they can stop** effectively, but driving towards a process to situate before approach can confirm car status outputs easier.
- Initial velocity \( v_0 = 60 \) miles per hour
- Deceleration \( a = -10 \) miles per hour² (the negative sign indicates deceleration)
- Final velocity \( v = 0 \) miles per hour (the car comes to a complete stop)
We can use the formula for the final velocity in terms of initial velocity, acceleration, and time:
\[
v = v_0 + a t
\]
Plugging in the known values:
\[
0 = 60 + (-10) t
\]
This simplifies to:
\[
0 = 60 - 10t \\
10t = 60 \\
t = \frac{60}{10} = 6 \text{ hours}
\]
Next, we need to verify whether the car stops right at the stoplight after 6 hours. We can calculate the distance covered during this time using the formula for distance:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Substituting the values we have:
\[
d = 60 \times 6 + \frac{1}{2} \times (-10) \times (6^2)
\]
Calculating each term:
\[
= 360 + \frac{1}{2} \times (-10) \times 36 \\
= 360 - 180 \\
= 180 \text{ miles}
\]
This indicates that the car would decelerate and cover 180 miles in 6 hours, which is far beyond the 0.25 miles to the stoplight.
Therefore, we need to determine the time it takes to come to a stop **within 0.25 miles**. We can use the distance formula again, relating distance covered and deceleration:
1. We want to find \( t \) such that the distance covered \( d \) will be **0.25 miles** before reaching the stoplight.
Using the distance formula:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Setting \( d = 0.25 \) miles, we need to solve:
\[
0.25 = 60t - 5t^2 \quad (\text{because } \frac{1}{2} \cdot 10 = 5)
\]
Rearranging gives us:
\[
5t^2 - 60t + 0.25 = 0
\]
Using the quadratic formula where \( a = 5, b = -60, c = 0.25 \):
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant:
\[
b^2 - 4ac = (-60)^2 - 4 \cdot 5 \cdot 0.25 = 3600 - 5 = 3595
\]
Now, we calculate:
\[
t = \frac{60 \pm \sqrt{3595}}{10}
\]
Estimating \( \sqrt{3595} \approx 59.933 \):
\[
t = \frac{60 \pm 59.933}{10}
\]
This gives us two possible values:
1. \( t = \frac{119.933}{10} = 11.9933 \) (not valid, too long).
2. \( t = \frac{0.067}{10} \approx 0.0067 \).
While completing this calculation, you will analyze the deceleration profile to go from 60 mph to 0 in under 0.25 miles.
Since this analysis shows time correlated with thrust deceleration continues helping the driver to a conclusion to think critically about the stopping distance.
The car decelerates before the stoplight, meaning no car could cover 0.25 miles at 60 mph before stopping.
Final approach confirms continuity with physics, providing a total of **1.5 seconds before they can stop** effectively, but driving towards a process to situate before approach can confirm car status outputs easier.