Time to travel a given distance and acceleration

Question

At 1200 hrs a car moving at $8 \dfrac{meters}{s}$is found accelerating $\dfrac{0.4 \;meters}{s^2}$.

When does it go another 100 meters? What is the easy way to find time for next 100 meters travel.

EDIT1:

One way is to use derived formula after verification from its quadratic equation solution:

$$ \boxed{ u= \frac{s}{t}-\frac12 at ;\quad v = \frac{s}{t} +\frac12 at \;} $$

and find physical basis of of choice of one from two solutions.

velocity, acceleration, distance

Answer 1

$$v(t) = v_0 + a_0\,t$$

By integrating this expression, considering that $x(0) = x_0$:

$$x(t)-x_0 = v_0\,t + \frac{a_0}{2}\,t^2$$

Thus,

$$\Delta x = v_0\,t + \frac{a_0}{2}\,t^2$$

$$100 = 8\,t + 0.2\,t^2 \Rightarrow t^2 + 40\,t - 500 = 0 \Rightarrow t=\frac{-40\pm\sqrt{40^2+4\cdot500}}{2} = -20\pm30$$

Since we want a future time in which this expression is satisfied, we take the positive answer $t=-20+30=10$. So, in 10 seconds it will move 100 meters.

Answer 2

We have: $$x(t)=x_0+v_0t+a_0\frac{t^2}{2}$$

Now solve:

$$100=8t+\frac{t^2}{2}0.4$$