A runner hopes to complete the 10,000-m run in less than 30.0 min. After running at constant speed for exactly 27.0 min, there are still 1200 m to go. The runner must then accelerate at$\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{2}}$for how many seconds in order to achieve the desired time?

Acceleration is a vector quantity. It is equal to the second-order derivative of displacement and the first-order derivative of velocity. Its magnitude at any instant is equal to the rate at which velocity changes.

Length of the track,$L=10,000\mathrm{m}$

Acceleration of the runner for the last 1200 m,$a=0.20{\mathrm{ms}}^{-2}$

Let the runner accelerate for time t.

The runner completes the initial distance (10,000 m–1200 m) at a constant speed in 27 minutes.

Let the speed be v, then.

$\begin{array}{c}v=\frac{10,000-1200}{27\times 60}\\ =5.43{\mathrm{ms}}^{-1}\end{array}$

Let the runner have an acceleration of a for time t and complete the race in 30 minutes. He covers 1200 m in 3 minutes, the equation for which is.

$1200=\left(vt+\frac{1}{2}a{t}^2\right)+\left(v+at\right)\left(3\times 60-t\right)$

Substituting the values in the above equation,

$1200=\left(5.43t+\frac{1}{2}\times 0.20t^2\right)+\left(5.43+0.20t\right)\left(3\times 60-t\right)$

The above equation can be re-written as.

$0.10t^2-36t+222.6=0$

The roots of the above quadratic equation are –366.08 and 6.08. As time cannot be negative, role="math" localid="1642846705371" $t=6.08\mathrm{s}$.

Thus, the runner must accelerate for 6.08 s.