A train is moving parallel and adjacent to a highway with a constant speed of 27 m/s. Ini- tially a car is 60 m behind the train, traveling in the same direction as the train at 41 m/s and accelerating at 5 m/s2.
What is the speed of the car just as it passes the train?
Answer in units of m/s
Dc = Vo*t + 0.5a*t^2 = 27t+60
41t + 0.5*5*t^2 = 27t + 60
41t -27t + 2.5t^2 = 60
14t + 2.5t^2 = 60
2.5t^2 + 14t - 60 = 0
Use Quadratic formula.
t = 2.84 s. To catch up.
Vc = Vo + a*t = 41 + 5*2.84 = 55.2 m/s.
To find the speed of the car just as it passes the train, we need to determine the time it takes for the car to catch up to the position of the train.
Let's start by finding the time it takes for the car to catch up to the train. We can use the equation of motion:
\[d = ut + \frac{1}{2}at^2\]
Where:
- \(d\) is the distance traveled by the car
- \(u\) is the initial velocity of the car
- \(a\) is the acceleration of the car
- \(t\) is the time taken
From the problem statement, we know that \(u = 41 \, \text{m/s}\), \(a = 5 \, \text{m/s}^2\), and \(d = 60 \, \text{m}\).
Plugging in these values into the equation of motion:
\[60 = 41t + \frac{1}{2} \cdot 5 \cdot t^2\]
Simplifying, let's rewrite this equation:
\[5t^2 + 41t - 120 = 0\]
Now, we can solve this quadratic equation for \(t\). Using the quadratic formula:
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
where \(a = 5\), \(b = 41\), and \(c = -120\).
Plugging in these values:
\[t = \frac{-41 \pm \sqrt{41^2 - 4 \cdot 5 \cdot -120}}{2 \cdot 5}\]
Simplifying further:
\[t = \frac{-41 \pm \sqrt{1681 + 2400}}{10}\]
\[t = \frac{-41 \pm \sqrt{4081}}{10}\]
Since time cannot be negative, we can ignore the negative value. Therefore, we have:
\[t = \frac{-41 + \sqrt{4081}}{10} \approx 2.056 \, \text{s}\]
Now that we have the time it takes for the car to catch up to the train, we can calculate the speed of the car just as it passes the train. We know that the speed of the train is constant at 27 m/s. Therefore, the speed of the car as it passes the train is also 27 m/s.
Hence, the speed of the car just as it passes the train is 27 m/s.