calculate the time taken for a car to cover a distance of 125m if the initial speed is 5m/s AMD it has a constant acceleration of 1.5m/s?
solve for t in
5t + 1.5/2 t^2 = 125
Does that kind of equation look familiar?
s = s0 + v0t + 1/2 at^2
To calculate the time taken for a car to cover a distance given the initial speed and constant acceleration, you can use the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
Where:
- \(s\) is the distance covered
- \(u\) is the initial velocity
- \(a\) is the constant acceleration
- \(t\) is the time taken
In this case, we have:
- \(s = 125\) m (distance covered)
- \(u = 5\) m/s (initial velocity)
- \(a = 1.5\) m/s² (constant acceleration)
Plugging in these values into the equation, we get:
\(125 = 5t + \frac{1}{2} \cdot 1.5t^2\)
Simplifying this equation, we have a quadratic equation:
\(1.5t^2 + 5t - 125 = 0\)
To solve this quadratic equation, we can use the quadratic formula:
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Where:
- \(a = 1.5\)
- \(b = 5\)
- \(c = -125\)
Plugging in these values into the quadratic formula, we get:
\(t = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1.5 \cdot -125}}{2 \cdot 1.5}\)
Simplifying further, we get:
\(t = \frac{-5 \pm \sqrt{25 + 750}}{3}\)
\(t = \frac{-5 \pm \sqrt{775}}{3}\)
Therefore, the time taken for the car to cover a distance of 125 m is approximately:
\(t \approx \frac{-5 + \sqrt{775}}{3}\) seconds and \(\frac{-5 - \sqrt{775}}{3}\) seconds.