Question

A sports car can accelerate uniformly to a speed of $162\text{km/h}$ in $5\text{s}$. Its maximum braking retardation is $6{\text{m/s}}^2$. The minimum time in which it can travel $1\text{km}$, starting from rest and ending at rest, in seconds is?

Answer 1

For the car to cover the distance in minimum time, it must continue to accelerate for as long as possible. If the car first constantly accelerates, and then immediately constantly decelerates, then the minimum time is achieved.

Firstly, we need to calculate the maximum acceleration of the car. We know that the car reaches the speed of $162\text{km/h}$ after $5\text{s}$ of acceleration. Let's assume $v$ to be the final speed of the car.
Then,
$v=162\text{km/h}=162\times \frac{5}{18}=45\text{m/s}$
So we know,
$v=45\text{m/s}$
$u=0\text{m/s}$
$t=5\text{s}$
By using $v=u+at$ we get,
$\Rightarrow 45=0+a\times 5$
$\Rightarrow a=9{\text{m/s}}^2$

Let's assume that to cover $1\text{km}$, the car takes time $t\text{s}$.
Also, let's assume while covering this distance the car accelerated for $t_0\text{s}$.
From $0\text{s}$ to $t_0\text{s}$, we know:
$v=v_0\text{m/s}$
$u=0\text{m/s}$
$a=9{\text{m/s}}^2$
$t=t_0\text{s}$
By using $v=u+at$ we get,
$\Rightarrow v_0=0+9t_0=9t_0$

From $t_0\text{s}$ to $t\text{s}$,
We know,
$v=0\text{m/s}$
$u=v_0\text{m/s}$
$a=-6{\text{m/s}}^2$
$t=(t-t_0)\text{s}$
By using $v=u+at$ we get,
$\Rightarrow 0=v_0-6(t-t_0)$
$\Rightarrow v_0=6(t-t_0)$
But $v_0=9t_0$
$\Rightarrow 9t_0=6(t-t_0)$
$\Rightarrow (t-t_0)=t_0\times \frac{3}{2}$
$\Rightarrow t=t_0\times \frac{5}{2}$

Applying $s=ut+\frac{1}{2}a{t}^2$ from $0\text{s}$ to $t_0\text{s}$,
$s_1=0+\frac{1}{2}\times 9\times (t_0{)}^2$
$s_1=\frac{9}{2}{t}_0^2$
Applying $s=ut+\frac{1}{2}a{t}^2$ from $t_0\text{s}$ to $t\text{s}$,
$s_2=\left(9t_0\right)\times (t-t_0)+\frac{1}{2}\times (-6)\times (t-t_0{)}^2$
$s_2=\left(9t_0\right)\times \frac{3}{2}{t}_0-3(\frac{3}{2}{t}_0{)}^2$
$s_2=\frac{3}{2}\times \frac{9}{2}{t}_0^2$

We know,
$s_1+s_2=1000\text{m}=(1+\frac{3}{2})\times \frac{9}{2}{t}_0^2$
$t_0^2=\frac{4}{45}\times 1000$
$t_0=9.42809\text{s}$
$t=\frac{5}{2}\times t_0=23.57022\text{s}$