Question
What is the time it takes for a car to travel 250 meters with an initial velocity of 15 m/s and an acceleration of 2 m/s^2?
Answers
GPT-4o mini
To find the time it takes for a car to travel a distance of 250 meters, given an initial velocity (\(u\)) of 15 m/s and an acceleration (\(a\)) of 2 m/s², we can use the kinematic equation:
\[
s = ut + \frac{1}{2} a t^2
\]
where:
- \(s\) = distance traveled (250 meters)
- \(u\) = initial velocity (15 m/s)
- \(a\) = acceleration (2 m/s²)
- \(t\) = time (which we need to find)
Substituting the values into the equation:
\[
250 = 15t + \frac{1}{2} \cdot 2 \cdot t^2
\]
This simplifies to:
\[
250 = 15t + t^2
\]
Rearranging gives us a quadratic equation:
\[
t^2 + 15t - 250 = 0
\]
Now we can apply the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our equation, \(a = 1\), \(b = 15\), and \(c = -250\).
Calculating the discriminant:
\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot (-250) = 225 + 1000 = 1225
\]
Now find the square root of the discriminant:
\[
\sqrt{1225} = 35
\]
Substituting back into the quadratic formula:
\[
t = \frac{-15 \pm 35}{2 \cdot 1}
\]
This gives us two possible solutions:
\[
t = \frac{20}{2} = 10 \quad \text{(valid solution)}
\]
\[
t = \frac{-50}{2} = -25 \quad \text{(not valid as time cannot be negative)}
\]
Thus, the time it takes for the car to travel 250 meters is:
\[
\boxed{10 \text{ seconds}}
\]
\[
s = ut + \frac{1}{2} a t^2
\]
where:
- \(s\) = distance traveled (250 meters)
- \(u\) = initial velocity (15 m/s)
- \(a\) = acceleration (2 m/s²)
- \(t\) = time (which we need to find)
Substituting the values into the equation:
\[
250 = 15t + \frac{1}{2} \cdot 2 \cdot t^2
\]
This simplifies to:
\[
250 = 15t + t^2
\]
Rearranging gives us a quadratic equation:
\[
t^2 + 15t - 250 = 0
\]
Now we can apply the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our equation, \(a = 1\), \(b = 15\), and \(c = -250\).
Calculating the discriminant:
\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot (-250) = 225 + 1000 = 1225
\]
Now find the square root of the discriminant:
\[
\sqrt{1225} = 35
\]
Substituting back into the quadratic formula:
\[
t = \frac{-15 \pm 35}{2 \cdot 1}
\]
This gives us two possible solutions:
\[
t = \frac{20}{2} = 10 \quad \text{(valid solution)}
\]
\[
t = \frac{-50}{2} = -25 \quad \text{(not valid as time cannot be negative)}
\]
Thus, the time it takes for the car to travel 250 meters is:
\[
\boxed{10 \text{ seconds}}
\]