Calculate the time taken for atrain to travel 1000m.if its initial speed 30m/s and has constant acceralation of 0.1m/s^2.

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To calculate the time taken for the train to travel 1000 meters, we can use the equation of motion for uniformly accelerated linear motion:

\[s = ut + 0.5at^2\]

where:
- \(s\) is the distance traveled,
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(t\) is the time.

Given:
\(s = 1000\) m (distance)
\(u = 30\) m/s (initial speed)
\(a = 0.1\) m/s² (acceleration)

Let's substitute these values into the equation and solve for \(t\):

\[1000 = (30)t + 0.5(0.1)t^2\]

Simplifying the equation:

\[1000 = 30t + 0.05t^2\]

Rearranging the equation and making it a quadratic equation:

\[0.05t^2 + 30t - 1000 = 0\]

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's solve it using the quadratic formula:

\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For the quadratic equation \(0.05t^2 + 30t - 1000 = 0\), we have:
\(a = 0.05\),
\(b = 30\),
\(c = -1000\).

Substituting these values into the quadratic formula:

\[t = \frac{-30 \pm \sqrt{30^2 - 4(0.05)(-1000)}}{2(0.05)}\]

Simplifying further:

\[t = \frac{-30 \pm \sqrt{900 + 200}}{0.1}\]
\[t = \frac{-30 \pm \sqrt{1100}}{0.1}\]

Now, calculate the two possible values of \(t\):

\[t_1 = \frac{-30 + \sqrt{1100}}{0.1}\]
\[t_2 = \frac{-30 - \sqrt{1100}}{0.1}\]

By calculating these values, we find that \(t_1 \approx 14.77\) seconds and \(t_2 \approx -174.77\) seconds. Since time cannot be negative in this context, the train takes approximately 14.77 seconds to travel 1000 meters.